Q1 JEE Main 2026 · 2 APR · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
An electromagnetic wave travels in free space along the x-direction. At a particular point in space and time, B ⃗ = 2 × 1 0 − 7 j ^ \vec{B} = 2\times 10^{-7}\hat{j} B = 2 × 1 0 − 7 j ^ T is associated with this wave. The value of corresponding electric field E ⃗ \vec{E} E at this point is _______ V/m.
(a) 60 k ^ 60\,\hat{k} 60 k ^ (b) − 60 k ^ -60\,\hat{k} − 60 k ^ (c) 30 k ^ 30\,\hat{k} 30 k ^ (d) − 600 k ^ -600\,\hat{k} − 600 k ^ Show answer and solution
Answer: (b)
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Q2 JEE Main 2026 · 6 APR · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
For an electromagnetic wave propagating through vacuum, k ⃗ , E ⃗ \vec{k}, \vec{E} k , E and ω \omega ω represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by:
(a) E ⃗ × k ⃗ ω \frac{\vec{E}\times\vec{k}}{\omega} ω E × k (b) k ⃗ × E ⃗ ω \frac{\vec{k}\times\vec{E}}{\omega} ω k × E (c) ω ( E ⃗ × k ⃗ ) \omega(\vec{E}\times\vec{k}) ω ( E × k ) (d) ω ( k ⃗ × E ⃗ ) \omega(\vec{k}\times\vec{E}) ω ( k × E ) Show answer and solution
Answer: (b)
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Q3 JEE Main 2026 · 4 APR · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
A magnetic field vector in an electromagnetic wave is represented by B ⃗ = B 0 sin ( 2 π ν t − 2 π x λ ) j ^ \vec{B} = B_0\sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{j} B = B 0 sin ( 2 π ν t − λ 2 π x ) j ^ . Its associated electric field vector is ______ .
(a) E ⃗ = − ν λ B 0 sin ( 2 π ν t − 2 π x λ ) k ^ \vec{E} = -\nu\lambda B_0\sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{k} E = − ν λ B 0 sin ( 2 π ν t − λ 2 π x ) k ^ (b) E ⃗ = − ν λ B 0 sin ( 2 π ν t − 2 π x λ ) i ^ \vec{E} = -\nu\lambda B_0\sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{i} E = − ν λ B 0 sin ( 2 π ν t − λ 2 π x ) i ^ (c) E ⃗ = ν λ B 0 sin ( 2 π ν t − 2 π x λ ) k ^ \vec{E} = \nu\lambda B_0\sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{k} E = ν λ B 0 sin ( 2 π ν t − λ 2 π x ) k ^ (d) E ⃗ = ν λ B 0 sin ( 2 π ν t − 2 π x λ ) i ^ \vec{E} = \nu\lambda B_0\sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{i} E = ν λ B 0 sin ( 2 π ν t − λ 2 π x ) i ^ Show answer and solution
Answer: (a)
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Q4 JEE Main 2026 · 28 JAN · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
A plane electromagnetic wave is moving in free space with velocity c = 3 × 1 0 8 c = 3\times 10^{8} c = 3 × 1 0 8 m/s and its electric field is given as E ⃗ = 54 sin ( k z − ω t ) j ^ \vec{E} = 54\sin(kz - \omega t)\,\hat{j} E = 54 sin ( k z − ω t ) j ^ V/m, where j ^ \hat{j} j ^ is the unit vector along y-axis. The magnetic field vector B ⃗ \vec{B} B of the wave is:
(a) − 1.8 × 1 0 − 7 sin ( k z − ω t ) i ^ -1.8\times 10^{-7}\sin(kz - \omega t)\,\hat{i} − 1.8 × 1 0 − 7 sin ( k z − ω t ) i ^ T(b) + 1.8 × 1 0 − 7 sin ( k z − ω t ) i ^ +1.8\times 10^{-7}\sin(kz - \omega t)\,\hat{i} + 1.8 × 1 0 − 7 sin ( k z − ω t ) i ^ T(c) 1.4 × 1 0 − 7 sin ( k z − ω t ) k ^ 1.4\times 10^{-7}\sin(kz - \omega t)\,\hat{k} 1.4 × 1 0 − 7 sin ( k z − ω t ) k ^ T(d) 1.4 × 1 0 − 7 sin ( k z − ω t ) i ^ 1.4\times 10^{-7}\sin(kz - \omega t)\,\hat{i} 1.4 × 1 0 − 7 sin ( k z − ω t ) i ^ TShow answer and solution
Answer: (a)
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Q5 JEE Main 2026 · 21 JAN · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field in a plane electromagnetic wave is given by:
E y = 69 sin [ 0.6 × 1 0 3 x − 1.8 × 1 0 11 t ] V/m. E_y = 69\sin\left[0.6\times 10^{3}x - 1.8\times 10^{11}t\right]\text{ V/m.} E y = 69 sin [ 0.6 × 1 0 3 x − 1.8 × 1 0 11 t ] V/m.
The expression for magnetic field associated with this electromagnetic wave is ______ T.
(a) B z = 2.3 × 1 0 − 7 sin [ 0.6 × 1 0 3 x − 1.8 × 1 0 11 t ] B_z = 2.3\times 10^{-7}\sin\left[0.6\times 10^{3}x - 1.8\times 10^{11}t\right] B z = 2.3 × 1 0 − 7 sin [ 0.6 × 1 0 3 x − 1.8 × 1 0 11 t ] (b) B z = 2.3 × 1 0 − 7 sin [ 0.6 × 1 0 3 x + 1.8 × 1 0 11 t ] B_z = 2.3\times 10^{-7}\sin\left[0.6\times 10^{3}x + 1.8\times 10^{11}t\right] B z = 2.3 × 1 0 − 7 sin [ 0.6 × 1 0 3 x + 1.8 × 1 0 11 t ] (c) B y = 2.3 × 1 0 − 7 sin [ 0.6 × 1 0 3 x − 1.8 × 1 0 11 t ] B_y = 2.3\times 10^{-7}\sin\left[0.6\times 10^{3}x - 1.8\times 10^{11}t\right] B y = 2.3 × 1 0 − 7 sin [ 0.6 × 1 0 3 x − 1.8 × 1 0 11 t ] (d) B y = 69 sin [ 0.6 × 1 0 3 x + 1.8 × 1 0 11 t ] B_y = 69\sin\left[0.6\times 10^{3}x + 1.8\times 10^{11}t\right] B y = 69 sin [ 0.6 × 1 0 3 x + 1.8 × 1 0 11 t ] Show answer and solution
Answer: (a)
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Q6 JEE Main 2026 · 2 APR · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Force on a charge
An electromagnetic wave travelling in x-direction is described by field equation
E y = 300 sin ω ( t − x c ) E_y = 300 \sin \omega \left(t - \frac{x}{c}\right) E y = 300 sin ω ( t − c x ) .
If the electron is restricted to move in y-direction only with speed of 1.5 × 1 0 6 1.5 \times 10^{6} 1.5 × 1 0 6 m/s then ratio of maximum electric and magnetic forces acting on the electron is ______.
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Answer: (a)
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Q7 JEE Main 2025 · 29 JAN · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
A plane electromagnetic wave propagates along the + x direction in free space. The components of the electric field, E ⃗ \vec{E} E and magnetic field, B ⃗ \vec{B} B vectors associated with the wave in Cartesian frame are
(a) E x , B y E_x, B_y E x , B y (b) E y , B x E_y, B_x E y , B x (c) E y , B z E_y, B_z E y , B z (d) E z , B y E_z, B_y E z , B y Show answer and solution
Answer: (c)
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Q8 NEET 2025 Easy ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field in a plane electromagnetic wave is given by E z = 60 cos ( 5 x + 1.5 × 1 0 9 t ) V/m E_z = 60\cos\left(5x + 1.5\times 10^{9}t\right)\text{ V/m} E z = 60 cos ( 5 x + 1.5 × 1 0 9 t ) V/m Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
(a) B z = 60 cos ( 5 x + 1.5 × 1 0 9 t ) T B_z = 60\cos\left(5x + 1.5\times 10^{9}t\right)\text{ T} B z = 60 cos ( 5 x + 1.5 × 1 0 9 t ) T (b) B y = 60 sin ( 5 x + 1.5 × 1 0 9 t ) T B_y = 60\sin\left(5x + 1.5\times 10^{9}t\right)\text{ T} B y = 60 sin ( 5 x + 1.5 × 1 0 9 t ) T (c) B y = 2 × 1 0 − 7 cos ( 5 x + 1.5 × 1 0 9 t ) T B_y = 2\times 10^{-7}\cos\left(5x + 1.5\times 10^{9}t\right)\text{ T} B y = 2 × 1 0 − 7 cos ( 5 x + 1.5 × 1 0 9 t ) T (d) B x = 2 × 1 0 − 7 cos ( 5 x + 1.5 × 1 0 9 t ) T B_x = 2\times 10^{-7}\cos\left(5x + 1.5\times 10^{9}t\right)\text{ T} B x = 2 × 1 0 − 7 cos ( 5 x + 1.5 × 1 0 9 t ) T Show answer and solution
Answer: (c)
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Q9 JEE Main 2025 · 23 JAN · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
A plane electromagnetic wave of frequency 20 MHz travels in free space along the + x +x + x direction. At a particular point in space and time, the electric field vector of the wave is E y = 9.3 V m − 1 E_y = 9.3\text{ V m}^{-1} E y = 9.3 V m − 1 . Then, the magnetic field vector of the wave at that point is
(a) B z = 1.55 × 1 0 − 8 T B_z = 1.55\times 10^{-8}\text{ T} B z = 1.55 × 1 0 − 8 T (b) B z = 6.2 × 1 0 − 8 T B_z = 6.2\times 10^{-8}\text{ T} B z = 6.2 × 1 0 − 8 T (c) B z = 3.1 × 1 0 − 8 T B_z = 3.1\times 10^{-8}\text{ T} B z = 3.1 × 1 0 − 8 T (d) B z = 9.3 × 1 0 − 8 T B_z = 9.3\times 10^{-8}\text{ T} B z = 9.3 × 1 0 − 8 T Show answer and solution
Answer: (c)
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Q10 JEE Main 2025 · 28 JAN · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The magnetic field of an E.M. wave is given by B ⃗ = ( 3 2 i ^ + 1 2 j ^ ) 30 sin [ ω ( t − z c ) ] \vec{B} = \left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right)30\sin\left[\omega\left(t - \frac{z}{c}\right)\right] B = ( 2 3 i ^ + 2 1 j ^ ) 30 sin [ ω ( t − c z ) ] (S.I. Units). The corresponding electric field in S.I. units is:
(a) E ⃗ = ( 1 2 i ^ + 3 2 j ^ ) 30 c sin [ ω ( t + z c ) ] \vec{E} = \left(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right)30c\sin\left[\omega\left(t + \frac{z}{c}\right)\right] E = ( 2 1 i ^ + 2 3 j ^ ) 30 c sin [ ω ( t + c z ) ] (b) E ⃗ = ( 1 2 i ^ − 3 2 j ^ ) 30 c sin [ ω ( t − z c ) ] \vec{E} = \left(\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\right)30c\sin\left[\omega\left(t - \frac{z}{c}\right)\right] E = ( 2 1 i ^ − 2 3 j ^ ) 30 c sin [ ω ( t − c z ) ] (c) E ⃗ = ( 3 2 i ^ − 1 2 j ^ ) 30 c sin [ ω ( t + z c ) ] \vec{E} = \left(\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}\right)30c\sin\left[\omega\left(t + \frac{z}{c}\right)\right] E = ( 2 3 i ^ − 2 1 j ^ ) 30 c sin [ ω ( t + c z ) ] (d) E ⃗ = ( 3 4 i ^ + 1 4 j ^ ) 30 c cos [ ω ( t − z c ) ] \vec{E} = \left(\frac{3}{4}\hat{i} + \frac{1}{4}\hat{j}\right)30c\cos\left[\omega\left(t - \frac{z}{c}\right)\right] E = ( 4 3 i ^ + 4 1 j ^ ) 30 c cos [ ω ( t − c z ) ] Show answer and solution
Answer: (b)
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Q11 JEE Main 2025 · 23 JAN · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field of an electromagnetic wave in free space is
E ⃗ = 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( 4 i ^ − 3 j ^ ) N/C \vec{E} = 57\cos\left[7.5\times 10^{6}t - 5\times 10^{-3}(3x + 4y)\right](4\hat{i} - 3\hat{j})\text{ N/C} E = 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( 4 i ^ − 3 j ^ ) N/C . The associated magnetic field in Tesla is
(a) B ⃗ = 57 3 × 1 0 8 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( k ^ ) \vec{B} = \frac{57}{3\times 10^{8}}\cos\left[7.5\times 10^{6}t - 5\times 10^{-3}(3x + 4y)\right](\hat{k}) B = 3 × 1 0 8 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( k ^ ) (b) B ⃗ = 57 3 × 1 0 8 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( 5 k ^ ) \vec{B} = \frac{57}{3\times 10^{8}}\cos\left[7.5\times 10^{6}t - 5\times 10^{-3}(3x + 4y)\right](5\hat{k}) B = 3 × 1 0 8 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( 5 k ^ ) (c) B ⃗ = − 57 3 × 1 0 8 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( k ^ ) \vec{B} = -\frac{57}{3\times 10^{8}}\cos\left[7.5\times 10^{6}t - 5\times 10^{-3}(3x + 4y)\right](\hat{k}) B = − 3 × 1 0 8 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( k ^ ) (d) B ⃗ = − 57 3 × 1 0 8 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( 5 k ^ ) \vec{B} = -\frac{57}{3\times 10^{8}}\cos\left[7.5\times 10^{6}t - 5\times 10^{-3}(3x + 4y)\right](5\hat{k}) B = − 3 × 1 0 8 57 cos [ 7.5 × 1 0 6 t − 5 × 1 0 − 3 ( 3 x + 4 y ) ] ( 5 k ^ ) Show answer and solution
Answer: (d)
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Q12 NEET 2024 Easy ☆
Nature of EM Waves & Wave Equation · Nature of EM waves
The property which is not of an electromagnetic wave travelling in free space is that:
(a) They are transverse in nature (b) The energy density in electric field is equal to energy density in magnetic field (c) They travel with a speed equal to 1 μ 0 ε 0 \frac{1}{\sqrt{\mu_0\varepsilon_0}} μ 0 ε 0 1 (d) They originate from charges moving with uniform speed Show answer and solution
Answer: (d)
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Q13 JEE Main 2024 · 1 FEB · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Reading the wave equation
If frequency of electromagnetic wave is 60 MHz 60\text{ MHz} 60 MHz and it travels in air along z z z direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in m \text{m} m ) is:
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Answer: (b)
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Q14 JEE Main 2024 · 29 JAN · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
A plane electromagnetic wave of frequency 35 MHz 35\text{ MHz} 35 MHz travels in free space along the x x x -direction. At a particular point (in space and time) E ⃗ = 9.6 j ^ V/m \vec{E} = 9.6\hat{j}\text{ V/m} E = 9.6 j ^ V/m . The value of magnetic field at this point is:
(a) 9.6 j ^ T 9.6\hat{j}\text{ T} 9.6 j ^ T (b) 3.2 × 1 0 − 8 i ^ T 3.2\times 10^{-8}\hat{i}\text{ T} 3.2 × 1 0 − 8 i ^ T (c) 9.6 × 1 0 − 8 k ^ T 9.6\times 10^{-8}\hat{k}\text{ T} 9.6 × 1 0 − 8 k ^ T (d) 3.2 × 1 0 − 8 k ^ T 3.2\times 10^{-8}\hat{k}\text{ T} 3.2 × 1 0 − 8 k ^ T Show answer and solution
Answer: (d)
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Q15 JEE Main 2024 · 9 APR · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The magnetic field in a plane electromagnetic wave is
B y = ( 3.5 × 1 0 − 7 ) sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) T B_y = \left(3.5\times 10^{-7}\right)\sin\left(1.5\times 10^{3}x + 0.5\times 10^{11}t\right)\text{ T} B y = ( 3.5 × 1 0 − 7 ) sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) T . The corresponding electric field will be:
(a) E z = 105 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 E_z = 105\sin\left(1.5\times 10^{3}x + 0.5\times 10^{11}t\right)\text{ V m}^{-1} E z = 105 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 (b) E y = 10.5 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 E_y = 10.5\sin\left(1.5\times 10^{3}x + 0.5\times 10^{11}t\right)\text{ V m}^{-1} E y = 10.5 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 (c) E y = 1.17 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 E_y = 1.17\sin\left(1.5\times 10^{3}x + 0.5\times 10^{11}t\right)\text{ V m}^{-1} E y = 1.17 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 (d) E z = 1.17 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 E_z = 1.17\sin\left(1.5\times 10^{3}x + 0.5\times 10^{11}t\right)\text{ V m}^{-1} E z = 1.17 sin ( 1.5 × 1 0 3 x + 0.5 × 1 0 11 t ) V m − 1 Show answer and solution
Answer: (a)
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Q16 JEE Main 2024 · 9 APR · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
A plane EM wave is propagating along x x x direction. It has a wavelength of 4 mm 4\text{ mm} 4 mm . If electric field is in y y y direction with the maximum magnitude of 60 V m − 1 60\text{ V m}^{-1} 60 V m − 1 , the equation for magnetic field is:
(a) B z = 2 × 1 0 − 7 sin [ π 2 ( x − 3 × 1 0 8 t ) ] k ^ T B_z = 2\times 10^{-7}\sin\left[\frac{\pi}{2}\left(x - 3\times 10^{8}t\right)\right]\hat{k}\text{ T} B z = 2 × 1 0 − 7 sin [ 2 π ( x − 3 × 1 0 8 t ) ] k ^ T (b) B z = 2 × 1 0 − 7 sin [ π 2 × 1 0 3 ( x − 3 × 1 0 8 t ) ] k ^ T B_z = 2\times 10^{-7}\sin\left[\frac{\pi}{2}\times 10^{3}\left(x - 3\times 10^{8}t\right)\right]\hat{k}\text{ T} B z = 2 × 1 0 − 7 sin [ 2 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] k ^ T (c) B z = 60 sin [ π 2 ( x − 3 × 1 0 8 t ) ] k ^ T B_z = 60\sin\left[\frac{\pi}{2}\left(x - 3\times 10^{8}t\right)\right]\hat{k}\text{ T} B z = 60 sin [ 2 π ( x − 3 × 1 0 8 t ) ] k ^ T (d) B x = 60 sin [ π 2 ( x − 3 × 1 0 8 t ) ] i ^ T B_x = 60\sin\left[\frac{\pi}{2}\left(x - 3\times 10^{8}t\right)\right]\hat{i}\text{ T} B x = 60 sin [ 2 π ( x − 3 × 1 0 8 t ) ] i ^ T Show answer and solution
Answer: (b)
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Q17 JEE Main 2024 · 4 APR · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field in an electromagnetic wave is given by E ⃗ = i ^ 40 cos ω ( t − z / c ) NC − 1 \vec{E} = \hat{i}\,40\cos\omega(t - z/c)\text{ NC}^{-1} E = i ^ 40 cos ω ( t − z / c ) NC − 1 . The magnetic field induction of this wave is (in SI unit):
(a) B ⃗ = j ^ 40 c cos ω ( t − z / c ) \vec{B} = \hat{j}\frac{40}{c}\cos\omega(t - z/c) B = j ^ c 40 cos ω ( t − z / c ) (b) B ⃗ = i ^ 40 c cos ω ( t − z / c ) \vec{B} = \hat{i}\frac{40}{c}\cos\omega(t - z/c) B = i ^ c 40 cos ω ( t − z / c ) (c) B ⃗ = j ^ 40 cos ω ( t − z / c ) \vec{B} = \hat{j}\,40\cos\omega(t - z/c) B = j ^ 40 cos ω ( t − z / c ) (d) B ⃗ = k ^ 40 c cos ω ( t − z / c ) \vec{B} = \hat{k}\frac{40}{c}\cos\omega(t - z/c) B = k ^ c 40 cos ω ( t − z / c ) Show answer and solution
Answer: (a)
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Q18 JEE Main 2024 · 30 JAN · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field of an electromagnetic wave in free space is represented as E ⃗ = E 0 cos ( ω t − k z ) i ^ \vec{E} = E_0\cos(\omega t - kz)\hat{i} E = E 0 cos ( ω t − k z ) i ^ . The corresponding magnetic induction vector will be:
(a) B ⃗ = E 0 C cos ( ω t + k z ) j ^ \vec{B} = E_0 C\cos(\omega t + kz)\hat{j} B = E 0 C cos ( ω t + k z ) j ^ (b) B ⃗ = E 0 C cos ( ω t − k z ) j ^ \vec{B} = \frac{E_0}{C}\cos(\omega t - kz)\hat{j} B = C E 0 cos ( ω t − k z ) j ^ (c) B ⃗ = E 0 C cos ( ω t − k z ) j ^ \vec{B} = E_0 C\cos(\omega t - kz)\hat{j} B = E 0 C cos ( ω t − k z ) j ^ (d) B ⃗ = E 0 C cos ( ω t + k z ) j ^ \vec{B} = \frac{E_0}{C}\cos(\omega t + kz)\hat{j} B = C E 0 cos ( ω t + k z ) j ^ Show answer and solution
Answer: (b)
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Q19 MHT CET 2023 · 13 MAY · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Nature of EM waves
Select the 'WRONG' statement out of the following.
(a) Electromagnetic waves do not require any medium for their propagation. (b) Electromagnetic waves can travel through vacuum with speed of light (c) Material medium is necessary for propagation of electromagnetic waves. (d) Electromagnetic waves are transverse in nature. Show answer and solution
Answer: (c)
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Q20 JEE Main 2023 · 29 JAN · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Nature of EM waves
Given below are two statements:
Statement I: Electromagnetic waves are not deflected by electric and magnetic field. Statement II: The amplitude of electric field and the magnetic field in electromagnetic waves are related to each other as E 0 = μ 0 ε 0 B 0 E_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}}\,B_0 E 0 = ε 0 μ 0 B 0 .
In the light of the above statements, choose the correct answer from the options given below:
(a) Both Statement I and Statement II are true (b) Statement I is true and Statement II is false (c) Both Statement I and Statement II are false (d) Statement I is false but Statement II is true Show answer and solution
Answer: (b)
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Q21 JEE Main 2023 · 13 APR · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
In an electromagnetic wave, at an instant and at particular position, the electric field is along the negative z z z -axis and magnetic field is along the positive x x x -axis. Then the direction of propagation of electromagnetic wave is:
(a) at 4 5 ∘ 45^\circ 4 5 ∘ angle from positive y-axis (b) positive y y y -axis (c) negative y y y -axis (d) positive z-axis Show answer and solution
Answer: (c)
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Q22 JEE Main 2023 · 25 JAN · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
An electromagnetic wave is transporting energy in the negative z z z direction. At a certain point and certain time the direction of electric field of the wave is along positive y y y direction. What will be the direction of the magnetic field of the wave at that point and instant?
(a) Negative direction of y y y (b) Positive direction of z z z (c) Positive direction of x x x (d) Negative direction x x x Show answer and solution
Answer: (c)
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Q23 JEE Main 2023 · 24 JAN · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
If E ⃗ \vec{E} E and K ⃗ \vec{K} K represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by:
(ω \omega ω - angular frequency):
(a) 1 ω ( K ⃗ × E ⃗ ) \frac{1}{\omega}\left(\vec{K}\times\vec{E}\right) ω 1 ( K × E ) (b) K ⃗ × E ⃗ \vec{K}\times\vec{E} K × E (c) ω ( K ⃗ × E ⃗ ) \omega\left(\vec{K}\times\vec{E}\right) ω ( K × E ) (d) ω ( E ⃗ × K ⃗ ) \omega\left(\vec{E}\times\vec{K}\right) ω ( E × K ) Show answer and solution
Answer: (a)
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Q24 NEET 2023 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0 × 1 0 10 Hz 2.0\times 10^{10}\text{ Hz} 2.0 × 1 0 10 Hz and amplitude 48 V m − 1 48\text{ V m}^{-1} 48 V m − 1 . Then the amplitude of oscillating magnetic field is: (Speed of light in free space = 3 × 1 0 8 m s − 1 = 3\times 10^{8}\text{ m s}^{-1} = 3 × 1 0 8 m s − 1 )
(a) 1.6 × 1 0 − 8 T 1.6\times 10^{-8}\text{ T} 1.6 × 1 0 − 8 T (b) 1.6 × 1 0 − 7 T 1.6\times 10^{-7}\text{ T} 1.6 × 1 0 − 7 T (c) 1.6 × 1 0 − 6 T 1.6\times 10^{-6}\text{ T} 1.6 × 1 0 − 6 T (d) 1.6 × 1 0 − 9 T 1.6\times 10^{-9}\text{ T} 1.6 × 1 0 − 9 T Show answer and solution
Answer: (b)
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Q25 JEE Main 2023 · 11 APR · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
A plane electromagnetic wave of frequency 20 MHz 20\text{ MHz} 20 MHz propagates in free space along x x x -direction. At a particular space and time, E ⃗ = 6.6 j ^ V/m \vec{E} = 6.6\hat{j}\text{ V/m} E = 6.6 j ^ V/m . What is B ⃗ \vec{B} B at this point?
(a) − 2.2 × 1 0 − 8 i ^ T -2.2\times 10^{-8}\hat{i}\text{ T} − 2.2 × 1 0 − 8 i ^ T (b) 2.2 × 1 0 − 8 i ^ T 2.2\times 10^{-8}\hat{i}\text{ T} 2.2 × 1 0 − 8 i ^ T (c) 2.2 × 1 0 − 8 k ^ T 2.2\times 10^{-8}\hat{k}\text{ T} 2.2 × 1 0 − 8 k ^ T (d) − 2.2 × 1 0 − 8 k ^ T -2.2\times 10^{-8}\hat{k}\text{ T} − 2.2 × 1 0 − 8 k ^ T Show answer and solution
Answer: (c)
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Q26 JEE Main 2023 · 10 APR · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
The amplitude of magnetic field in an electromagnetic wave propagating along y-axis is 6.0 × 1 0 − 7 T 6.0\times 10^{-7}\text{ T} 6.0 × 1 0 − 7 T . The maximum value of electric field in the electromagnetic wave is
(a) 6.0 × 1 0 − 7 V m − 1 6.0\times 10^{-7}\text{ V m}^{-1} 6.0 × 1 0 − 7 V m − 1 (b) 5 × 1 0 14 V m − 1 5\times 10^{14}\text{ V m}^{-1} 5 × 1 0 14 V m − 1 (c) 180 V m − 1 180\text{ V m}^{-1} 180 V m − 1 (d) 2 × 1 0 15 V m − 1 2\times 10^{15}\text{ V m}^{-1} 2 × 1 0 15 V m − 1 Show answer and solution
Answer: (c)
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Q27 JEE Main 2023 · 24 JAN · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by
E x = E 0 sin ( k z − ω t ) E_x = E_0\sin(kz - \omega t) E x = E 0 sin ( k z − ω t )
B y = B 0 sin ( k z − ω t ) B_y = B_0\sin(kz - \omega t) B y = B 0 sin ( k z − ω t )
Then the correct relation between E0 {}_0 0 and B0 {}_0 0 is given by
(a) E 0 B 0 = ω k E_0 B_0 = \omega k E 0 B 0 = ωk (b) E 0 = k B 0 E_0 = k B_0 E 0 = k B 0 (c) k E 0 = ω B 0 k E_0 = \omega B_0 k E 0 = ω B 0 (d) ω E 0 = k B 0 \omega E_0 = k B_0 ω E 0 = k B 0 Show answer and solution
Answer: (c)
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Q28 MHT CET 2023 · 9 MAY · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Force on a charge
A charge 'q q q ' moves with velocity 'v v v ' through electric (E) as well as magnetic field (B). Then the force acting on it is
(a) q ( v ⃗ × B ⃗ ) q(\vec{v}\times\vec{B}) q ( v × B ) (b) q ( B ⃗ × v ⃗ ) q(\vec{B}\times\vec{v}) q ( B × v ) (c) q ( E ⃗ × v ⃗ ) q(\vec{E}\times\vec{v}) q ( E × v ) (d) q E ⃗ + q ( v ⃗ × B ⃗ ) q\vec{E} + q(\vec{v}\times\vec{B}) q E + q ( v × B ) Show answer and solution
Answer: (d)
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Q29 NEET 2022 · PHASE 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
The ratio of the magnitude of the magnetic field and electric field intensity of a plane electromagnetic wave in free space of permeability μ 0 \mu_0 μ 0 and permittivity ε 0 \varepsilon_0 ε 0 is (Given that c - velocity of light in free space)
(a) μ 0 ε 0 c \frac{\sqrt{\mu_0\varepsilon_0}}{c} c μ 0 ε 0 (b) c (c) 1 c \frac{1}{c} c 1 (d) c μ 0 ε 0 \frac{c}{\sqrt{\mu_0\varepsilon_0}} μ 0 ε 0 c Show answer and solution
Answer: (c)
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Q30 JEE Main 2022 · 30 JUL · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
A plane electromagnetic wave propagating in free space has amplitude of electric field of 900 N C 900\ \frac{\text{N}}{\text{C}} 900 C N . The amplitude of magnetic field will be :__________
(a) 9 μ T 9\ \mu\text{T} 9 μ T (b) 27 μ T 27\ \mu\text{T} 27 μ T (c) 3 μ T 3\ \mu\text{T} 3 μ T (d) 270 GT 270\text{ GT} 270 GT Show answer and solution
Answer: (c)
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Q31 JEE Main 2022 · 26 JUL · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
The oscillating magnetic field in a plane electromagnetic wave is given by
B y = 5 × 1 0 − 6 sin 1000 π ( 5 x − 4 × 1 0 8 t ) T B_y = 5\times 10^{-6}\sin 1000\pi\left(5x - 4\times 10^{8}t\right)\text{ T} B y = 5 × 1 0 − 6 sin 1000 π ( 5 x − 4 × 1 0 8 t ) T . The amplitude of electric field will be:
(a) 15 × 1 0 2 V m − 1 15\times 10^{2}\text{ V m}^{-1} 15 × 1 0 2 V m − 1 (b) 5 × 1 0 − 6 V m − 1 5\times 10^{-6}\text{ V m}^{-1} 5 × 1 0 − 6 V m − 1 (c) 16 × 1 0 12 V m − 1 16\times 10^{12}\text{ V m}^{-1} 16 × 1 0 12 V m − 1 (d) 4 × 1 0 2 V m − 1 4\times 10^{2}\text{ V m}^{-1} 4 × 1 0 2 V m − 1 Show answer and solution
Answer: (d)
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Q32 JEE Main 2022 · 25 JUL · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
Light wave travelling in air along x-direction is given by E y = 540 sin π × 1 0 4 ( x − c t ) V m − 1 E_y = 540\sin\pi\times 10^{4}(x - ct)\text{ V m}^{-1} E y = 540 sin π × 1 0 4 ( x − c t ) V m − 1 . Then, the peak value of magnetic field of wave will be (Given c = 3 × \times × 108 ms− 1 {}^{-1} − 1 )
(a) 18 × \times × 10 − - − 7 T (b) 54 × \times × 10− - − 7 T (c) 54 × \times × 10− - − 8 T (d) 18 × \times × 10− - − 8 T Show answer and solution
Answer: (a)
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Q33 JEE Main 2022 · 30 JUN · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
An expression for oscillating electric field in a plane electromagnetic wave is given as Ez = 300 sin(5π × \pi\times π × 103x − - − 3π × \pi\times π × 1011t) Vm− 1 {}^{-1} − 1
Then, the value of magnetic field amplitude will be:
(Given: speed of light in Vacuum c = 3 × \times × 108 ms− 1 {}^{-1} − 1 )
(a) 1 × \times × 10− - − 6 T (b) 5 × \times × 10− - − 6 T (c) 18 × \times × 109 T (d) 21 × \times × 109 T Show answer and solution
Answer: (b)
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Q34 JEE Main 2022 · 27 JUL · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Nature of EM waves
Identify the correct statements from the following descriptions of various properties of electromagnetic waves.
(A) In a plane electromagnetic wave electric field and magnetic field must be perpendicular to each other and direction of propagation of wave should be along electric field or magnetic field. (B) The energy in electromagnetic wave is divided equally between electric and magnetic fields. (C) Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation of wave.
(D) The electric field, magnetic field and direction of propagation of wave must be perpendicular to each other.
(E) The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light.
Choose the most appropriate answer from the options given below:
(a) (D) only (b) (B) and (D) only (c) (B), (C) and (E) only (d) (A), (B) and (E) only Show answer and solution
Answer: (b)
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Q35 JEE Main 2022 · 26 JUL · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
The magnetic field of a plane electromagnetic wave is given by:
B ⃗ = 2 × 1 0 − 8 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) j ^ T \vec{B} = 2\times 10^{-8}\sin\left(0.5\times 10^{3}x + 1.5\times 10^{11}t\right)\hat{j}\text{ T} B = 2 × 1 0 − 8 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) j ^ T .
The amplitude of the electric field would be:
(a) 6 V m − 1 6\text{ V m}^{-1} 6 V m − 1 along x x x -axis(b) 3 V m − 1 3\text{ V m}^{-1} 3 V m − 1 along z z z -axis(c) 6 V m − 1 6\text{ V m}^{-1} 6 V m − 1 along z z z -axis(d) 2 × 1 0 − 8 V m − 1 2\times 10^{-8}\text{ V m}^{-1} 2 × 1 0 − 8 V m − 1 along z z z -axisShow answer and solution
Answer: (c)
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Q36 JEE Main 2022 · 26 JUN · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
If Electric field intensity of a uniform plane electromagnetic wave is given as
E = − 301.6 sin ( k z − ω t ) a ^ x + 452.4 sin ( k z − ω t ) a ^ y V m E = -301.6\sin(kz - \omega t)\hat{a}_x + 452.4\sin(kz - \omega t)\hat{a}_y\,\frac{\text{V}}{\text{m}} E = − 301.6 sin ( k z − ω t ) a ^ x + 452.4 sin ( k z − ω t ) a ^ y m V . Then magnetic intensity 'H' of this wave in Am− 1 {}^{-1} − 1 will be:
[Given: Speed of light in vacuum c = 3 × 1 0 8 c = 3\times 10^{8} c = 3 × 1 0 8 ms− 1 {}^{-1} − 1 , Permeability of vacuum μ 0 = 4 π × 1 0 − 7 \mu_0 = 4\pi\times 10^{-7} μ 0 = 4 π × 1 0 − 7 NA− 2 {}^{-2} − 2 ]
(a) + 0.8 sin ( k z − ω t ) a ^ y + 0.8 sin ( k z − ω t ) a ^ x +0.8\sin(kz - \omega t)\hat{a}_y + 0.8\sin(kz - \omega t)\hat{a}_x + 0.8 sin ( k z − ω t ) a ^ y + 0.8 sin ( k z − ω t ) a ^ x (b) + 1.0 × 1 0 − 6 sin ( k z − ω t ) a ^ y + 1.5 × 1 0 − 6 sin ( k z − ω t ) a ^ x +1.0\times 10^{-6}\sin(kz - \omega t)\hat{a}_y + 1.5\times 10^{-6}\sin(kz - \omega t)\hat{a}_x + 1.0 × 1 0 − 6 sin ( k z − ω t ) a ^ y + 1.5 × 1 0 − 6 sin ( k z − ω t ) a ^ x (c) − 0.8 sin ( k z − ω t ) a ^ y − 1.2 sin ( k z − ω t ) a ^ x -0.8\sin(kz - \omega t)\hat{a}_y - 1.2\sin(kz - \omega t)\hat{a}_x − 0.8 sin ( k z − ω t ) a ^ y − 1.2 sin ( k z − ω t ) a ^ x (d) − 1.0 × 1 0 − 6 sin ( k z − ω t ) a ^ y − 1.5 × 1 0 − 6 sin ( k z − ω t ) a ^ x -1.0\times 10^{-6}\sin(kz - \omega t)\hat{a}_y - 1.5\times 10^{-6}\sin(kz - \omega t)\hat{a}_x − 1.0 × 1 0 − 6 sin ( k z − ω t ) a ^ y − 1.5 × 1 0 − 6 sin ( k z − ω t ) a ^ x Show answer and solution
Answer: (c)
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Q37 NEET 2022 · PHASE 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The magnetic field of a plane electromagnetic wave is given by
B ⃗ = 3 × 1 0 − 8 cos ( 1.6 × 1 0 3 x + 48 × 1 0 10 t ) j ^ \vec{B} = 3\times 10^{-8}\cos(1.6\times 10^{3}x + 48\times 10^{10}t)\hat{j} B = 3 × 1 0 − 8 cos ( 1.6 × 1 0 3 x + 48 × 1 0 10 t ) j ^ , then the associated electric field will be:
(a) 9 cos ( 1.6 × 1 0 3 x + 48 × 1 0 10 t ) k ^ 9\cos(1.6\times 10^{3}x + 48\times 10^{10}t)\hat{k} 9 cos ( 1.6 × 1 0 3 x + 48 × 1 0 10 t ) k ^ V/m(b) 3 × 1 0 − 8 cos ( 1.6 × 1 0 3 x + 48 × 1 0 10 t ) i ^ 3\times 10^{-8}\cos(1.6\times 10^{3}x + 48\times 10^{10}t)\hat{i} 3 × 1 0 − 8 cos ( 1.6 × 1 0 3 x + 48 × 1 0 10 t ) i ^ V/m(c) 3 × 1 0 − 8 sin ( 1.6 × 1 0 3 x + 48 × 1 0 10 t ) i ^ 3\times 10^{-8}\sin(1.6\times 10^{3}x + 48\times 10^{10}t)\hat{i} 3 × 1 0 − 8 sin ( 1.6 × 1 0 3 x + 48 × 1 0 10 t ) i ^ V/m(d) 9 sin ( 1.6 × 1 0 3 x − 48 × 1 0 10 t ) k ^ 9\sin(1.6\times 10^{3}x - 48\times 10^{10}t)\hat{k} 9 sin ( 1.6 × 1 0 3 x − 48 × 1 0 10 t ) k ^ V/mShow answer and solution
Answer: (a)
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Q38 JEE Main 2022 · 28 JUN · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y- direction has maximum magnitude of 60 Vm− 1 {}^{-1} − 1 . Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum:
(a) E y = 60 sin [ π 4 × 1 0 3 ( x − 3 × 1 0 8 t ) ] j ^ V m − 1 B z = 2 sin [ π 4 × 1 0 3 ( x − 3 × 1 0 8 t ) ] k ^ T E_y = 60\sin\left[\frac{\pi}{4}\times 10^{3}(x - 3\times 10^{8}t)\right]\hat{j}\text{ V m}^{-1}\quad B_z = 2\sin\left[\frac{\pi}{4}\times 10^{3}(x - 3\times 10^{8}t)\right]\hat{k}\text{ T} E y = 60 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] j ^ V m − 1 B z = 2 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] k ^ T (b) E y = 60 sin [ π 4 × 1 0 3 ( x − 3 × 1 0 8 t ) ] j ^ V m − 1 B z = 2 × 1 0 − 7 sin [ π 4 × 1 0 3 ( x − 3 × 1 0 8 t ) ] k ^ T E_y = 60\sin\left[\frac{\pi}{4}\times 10^{3}(x - 3\times 10^{8}t)\right]\hat{j}\text{ V m}^{-1}\quad B_z = 2\times 10^{-7}\sin\left[\frac{\pi}{4}\times 10^{3}(x - 3\times 10^{8}t)\right]\hat{k}\text{ T} E y = 60 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] j ^ V m − 1 B z = 2 × 1 0 − 7 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] k ^ T (c) E y = 2 × 1 0 − 7 sin [ π 4 × 1 0 3 ( x − 3 × 1 0 8 t ) ] j ^ V m − 1 B z = 60 sin [ π 4 × 1 0 3 ( x − 3 × 1 0 8 t ) ] k ^ T E_y = 2\times 10^{-7}\sin\left[\frac{\pi}{4}\times 10^{3}(x - 3\times 10^{8}t)\right]\hat{j}\text{ V m}^{-1}\quad B_z = 60\sin\left[\frac{\pi}{4}\times 10^{3}(x - 3\times 10^{8}t)\right]\hat{k}\text{ T} E y = 2 × 1 0 − 7 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] j ^ V m − 1 B z = 60 sin [ 4 π × 1 0 3 ( x − 3 × 1 0 8 t ) ] k ^ T (d) E y = 2 × 1 0 − 7 sin [ π 4 × 1 0 4 ( x − 4 × 1 0 8 t ) ] j ^ V m − 1 B z = 60 sin [ π 4 × 1 0 4 ( x − 4 × 1 0 8 t ) ] k ^ T E_y = 2\times 10^{-7}\sin\left[\frac{\pi}{4}\times 10^{4}(x - 4\times 10^{8}t)\right]\hat{j}\text{ V m}^{-1}\quad B_z = 60\sin\left[\frac{\pi}{4}\times 10^{4}(x - 4\times 10^{8}t)\right]\hat{k}\text{ T} E y = 2 × 1 0 − 7 sin [ 4 π × 1 0 4 ( x − 4 × 1 0 8 t ) ] j ^ V m − 1 B z = 60 sin [ 4 π × 1 0 4 ( x − 4 × 1 0 8 t ) ] k ^ T Show answer and solution
Answer: (b)
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Q39 JEE Main 2022 · 27 JUL · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Force on a charge
A beam of light travelling along X X X -axis is described by the electric field E y = 900 sin ω ( t − x / c ) E_y = 900 \sin \omega (t - x/c) E y = 900 sin ω ( t − x / c ) . The ratio of electric force to magnetic force on a charge q q q moving along Y Y Y -axis with a speed of 3 × 1 0 7 ms − 1 3 \times 10^{7}\text{ ms}^{-1} 3 × 1 0 7 ms − 1 will be:
(Given speed of light = 3 × 1 0 8 ms − 1 = 3 \times 10^{8}\text{ ms}^{-1} = 3 × 1 0 8 ms − 1 )
(a) 1: 1 (b) 1: 10 (c) 10: 1 (d) 1: 2 Show answer and solution
Answer: (c)
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Q40 NEET 2021 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
For a plane electromagnetic wave propagating in x-direction, which one of the following combination gives the correct possible directions for electric field (E) and magnetic field (B) respectively?
(a) − j ^ + k ^ , − j ^ + k ^ -\hat{j}+\hat{k},\ -\hat{j}+\hat{k} − j ^ + k ^ , − j ^ + k ^ (b) j ^ + k ^ , j ^ + k ^ \hat{j}+\hat{k},\ \hat{j}+\hat{k} j ^ + k ^ , j ^ + k ^ (c) − j ^ + k ^ , − j ^ − k ^ -\hat{j}+\hat{k},\ -\hat{j}-\hat{k} − j ^ + k ^ , − j ^ − k ^ (d) j ^ + k ^ , − j ^ − k ^ \hat{j}+\hat{k},\ -\hat{j}-\hat{k} j ^ + k ^ , − j ^ − k ^ Show answer and solution
Answer: (c)
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Q41 JEE Main 2021 · 20 JUL · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
In an electromagnetic wave the electric field vector and magnetic field vector are given as E ⃗ = E 0 i ^ \vec{E} = E_0\hat{i} E = E 0 i ^ and B ⃗ = B 0 k ^ \vec{B} = B_0\hat{k} B = B 0 k ^ respectively. The direction of propagation of electromagnetic wave is along:
(a) ( k ^ ) \left(\hat{k}\right) ( k ^ ) (b) j ^ \hat{j} j ^ (c) ( − k ^ ) \left(-\hat{k}\right) ( − k ^ ) (d) ( − j ^ ) \left(-\hat{j}\right) ( − j ^ ) Show answer and solution
Answer: (d)
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Q42 JEE Main 2021 · 18 MAR · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
A plane electromagnetic wave propagating along y-direction can have the following pair of electric field ( E ⃗ ) \left(\vec{E}\right) ( E ) and magnetic field ( B ⃗ ) \left(\vec{B}\right) ( B ) components.
(a) Ex, Bz or Ez, Bx (b) Ex, By or Ey, Bx (c) Ey, By or Ez, Bz (d) Ey, Bx or Ex, By Show answer and solution
Answer: (a)
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Q43 JEE Main 2021 · 18 MAR · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
A plane electromagnetic wave of frequency 100 MHz is travelling in vacuum along the x-direction. At a particular point in space and time, B ⃗ = 2.0 × 1 0 − 8 k ^ T \vec{B} = 2.0\times 10^{-8}\hat{k}\text{ T} B = 2.0 × 1 0 − 8 k ^ T . (where, k ^ \hat{k} k ^ is unit vector along z-direction) What is E ⃗ \vec{E} E at this point?
(a) 0.6 j ^ \hat{j} j ^ V/m (b) 6.0 k ^ \hat{k} k ^ V/m (c) 6.0 j ^ \hat{j} j ^ V/m (d) 0.6 k ^ \hat{k} k ^ V/m Show answer and solution
Answer: (c)
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Q44 JEE Main 2021 · 16 MAR · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a particular point in space and time,
B ⃗ \vec{B} B = 8.0 × \times × 10− - − 8 z ^ \hat{z} z ^ T. The value of electric field at this point is:
(speed of light = 3 × \times × 108 ms− 1 {}^{-1} − 1 )
x ^ y ^ z ^ \hat{x}\ \hat{y}\ \hat{z} x ^ y ^ z ^ are unit vectors along x, y and z directions.
(a) 2.6 x ^ \hat{x} x ^ V/m (b) − - − 24 x ^ \hat{x} x ^ V/m(c) 24 x ^ \hat{x} x ^ V/m (d) − - − 2.6 y ^ \hat{y} y ^ V/mShow answer and solution
Answer: (b)
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Q45 JEE Main 2021 · 27 AUG · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
A plane electromagnetic wave with frequency of 30 MHz travels in free space. At particular point in space and time, electric field is 6 V/m. The magnetic field at this point will be x × \times × 10− - − 8 T. The value of x is ___________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 2
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Q46 JEE Main 2021 · 26 AUG · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Force on a charge
A light beam is described by E = 800 sin ω ( t − x c ) E = 800 \sin \omega \left(t - \frac{x}{c}\right) E = 800 sin ω ( t − c x ) . An electron is allowed to move normal to the propagation of light beam with a speed of 3 × \times × 107 ms− 1 {}^{-1} − 1 . What is the maximum magnetic force exerted on the electron?
(a) 1.28 × \times × 10− - − 18 N (b) 1.28 × \times × 10− - − 21 N (c) 12.8 × \times × 10− - − 17 N (d) 12.8 × \times × 10− - − 18 N Show answer and solution
Answer: (d)
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Q47 JEE Main 2021 · 25 JUL · SHIFT 1 Difficult ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
A linearly polarized electromagnetic wave in vacuum
E = 3.1 cos [ ( 1.8 ) z − ( 5.4 × 1 0 6 ) t ] i ^ N/C E = 3.1 \cos\left[(1.8)z - (5.4 \times 10^{6})t\right] \hat{i}\text{ N/C} E = 3.1 cos [ ( 1.8 ) z − ( 5.4 × 1 0 6 ) t ] i ^ N/C
is incident normally on a perfectly reflecting wall at z = a. Choose the correct option
(a) The wavelength is 5.4 m (b) The frequency of electromagnetic wave is 54 × \times × 104 Hz. (c) The transmitted wave will be 3.1 cos [ ( 1.8 ) z − ( 5.4 × 1 0 6 ) t ] i ^ N/C 3.1 \cos\left[(1.8)z - (5.4 \times 10^{6})t\right] \hat{i}\text{ N/C} 3.1 cos [ ( 1.8 ) z − ( 5.4 × 1 0 6 ) t ] i ^ N/C (d) The reflected wave will be 3.1 cos [ ( 1.8 ) z + ( 5.4 × 1 0 6 ) t ] i ^ N/C 3.1 \cos\left[(1.8)z + (5.4 \times 10^{6})t\right] \hat{i}\text{ N/C} 3.1 cos [ ( 1.8 ) z + ( 5.4 × 1 0 6 ) t ] i ^ N/C Show answer and solution
Answer: (d)
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Q48 JEE Main 2021 · 31 AUG · SHIFT 2 Difficult ☆
Nature of EM Waves & Wave Equation · Force on a charge
The magnetic field vector of an electromagnetic wave is given by B = B 0 i ^ + j ^ 2 cos ( k z − ω t ) B = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz - \omega t) B = B 0 2 i ^ + j ^ cos ( k z − ω t ) ; where i ^ , j ^ \hat{i}, \hat{j} i ^ , j ^ represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q1 of 4π \pi π coulomb and q2 of 2π \pi π coulomb located at ( 0 , 0 , π k ) \left(0, 0, \frac{\pi}{k}\right) ( 0 , 0 , k π ) and ( 0 , 0 , 3 π k ) \left(0, 0, \frac{3\pi}{k}\right) ( 0 , 0 , k 3 π ) , respectively, have the same velocity of 0.5 c i ^ \hat{i} i ^ , (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is :-
(a) 2 2 : 1 2\sqrt{2} : 1 2 2 : 1 (b) 1 : 2 1 : \sqrt{2} 1 : 2 (c) 2: 1 (d) 2 : 1 \sqrt{2} : 1 2 : 1 Show answer and solution
Answer: (c)
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Q49 JEE Main 2020 · 2 SEP · SHIFT 2 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by k ^ \hat{k} k ^ and 2 i ^ − 2 j ^ 2\hat{i}-2\hat{j} 2 i ^ − 2 j ^ , respectively. What is the unit vector along direction of propagation of the wave?
(a) 1 5 ( i ^ + 2 j ^ ) \frac{1}{\sqrt{5}}\left(\hat{i}+2\hat{j}\right) 5 1 ( i ^ + 2 j ^ ) (b) 1 5 ( 2 i ^ + j ^ ) \frac{1}{\sqrt{5}}\left(2\hat{i}+\hat{j}\right) 5 1 ( 2 i ^ + j ^ ) (c) 1 2 ( i ^ + j ^ ) \frac{1}{\sqrt{2}}\left(\hat{i}+\hat{j}\right) 2 1 ( i ^ + j ^ ) (d) 1 2 ( j ^ + k ^ ) \frac{1}{\sqrt{2}}\left(\hat{j}+\hat{k}\right) 2 1 ( j ^ + k ^ ) Show answer and solution
Answer: (c)
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Q50 NEET 2020 · PHASE 2 Easy ☆
Nature of EM Waves & Wave Equation · Reading the wave equation
The magnetic field in a plane electromagnetic wave is given by,
B y = 2 × 1 0 − 7 sin ( π × 1 0 3 x + 3 π × 1 0 11 t ) T B_y = 2\times 10^{-7}\sin(\pi\times 10^{3}x + 3\pi\times 10^{11}t)\text{ T} B y = 2 × 1 0 − 7 sin ( π × 1 0 3 x + 3 π × 1 0 11 t ) T
Calculate the wavelength.
(a) π × 1 0 − 3 m \pi\times 10^{-3}\text{ m} π × 1 0 − 3 m (b) π × 1 0 3 m \pi\times 10^{3}\text{ m} π × 1 0 3 m (c) 2 × 1 0 − 3 m 2\times 10^{-3}\text{ m} 2 × 1 0 − 3 m (d) 2 × 1 0 3 m 2\times 10^{3}\text{ m} 2 × 1 0 3 m Show answer and solution
Answer: (c)
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Q51 JEE Main 2020 · 6 SEP · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
For a plane electromagnetic wave, the magnetic field at a point x and time t is
B ⃗ ( x , t ) = [ 1.2 × 1 0 − 7 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) k ^ ] \vec{B}(x,t) = \left[1.2\times 10^{-7}\sin\left(0.5\times 10^{3}x + 1.5\times 10^{11}t\right)\hat{k}\right] B ( x , t ) = [ 1.2 × 1 0 − 7 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) k ^ ] T
The instantaneous electric field E ⃗ \vec{E} E corresponding to B ⃗ \vec{B} B is:
(speed of light c = 3 × 108 ms–1)
(a) E ⃗ ( x , t ) = [ 36 sin ( 1 × 1 0 3 x + 1.5 × 1 0 11 t ) i ^ ] V m \vec{E}(x,t) = \left[36\sin\left(1\times 10^{3}x + 1.5\times 10^{11}t\right)\hat{i}\right]\frac{\text{V}}{\text{m}} E ( x , t ) = [ 36 sin ( 1 × 1 0 3 x + 1.5 × 1 0 11 t ) i ^ ] m V (b) E ⃗ ( x , t ) = [ 36 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) k ^ ] V m \vec{E}(x,t) = \left[36\sin\left(0.5\times 10^{3}x + 1.5\times 10^{11}t\right)\hat{k}\right]\frac{\text{V}}{\text{m}} E ( x , t ) = [ 36 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) k ^ ] m V (c) E ⃗ ( x , t ) = [ 36 sin ( 1 × 1 0 3 x + 0.5 × 1 0 11 t ) j ^ ] V m \vec{E}(x,t) = \left[36\sin\left(1\times 10^{3}x + 0.5\times 10^{11}t\right)\hat{j}\right]\frac{\text{V}}{\text{m}} E ( x , t ) = [ 36 sin ( 1 × 1 0 3 x + 0.5 × 1 0 11 t ) j ^ ] m V (d) E ⃗ ( x , t ) = [ − 36 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) j ^ ] V m \vec{E}(x,t) = \left[-36\sin\left(0.5\times 10^{3}x + 1.5\times 10^{11}t\right)\hat{j}\right]\frac{\text{V}}{\text{m}} E ( x , t ) = [ − 36 sin ( 0.5 × 1 0 3 x + 1.5 × 1 0 11 t ) j ^ ] m V Show answer and solution
Answer: (d)
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Q52 JEE Main 2020 · 4 SEP · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field of a plane electromagnetic wave is given by
E ⃗ = E 0 ( x ^ + y ^ ) sin ( k z − ω t ) \vec{E} = E_0\left(\hat{x} + \hat{y}\right)\sin(kz - \omega t) E = E 0 ( x ^ + y ^ ) sin ( k z − ω t )
Its magnetic field will be given by:
(a) E 0 c ( x ^ + y ^ ) sin ( k z − ω t ) \frac{E_0}{c}\left(\hat{x} + \hat{y}\right)\sin(kz - \omega t) c E 0 ( x ^ + y ^ ) sin ( k z − ω t ) (b) E 0 c ( x ^ − y ^ ) sin ( k z − ω t ) \frac{E_0}{c}\left(\hat{x} - \hat{y}\right)\sin(kz - \omega t) c E 0 ( x ^ − y ^ ) sin ( k z − ω t ) (c) E 0 c ( x ^ − y ^ ) cos ( k z − ω t ) \frac{E_0}{c}\left(\hat{x} - \hat{y}\right)\cos(kz - \omega t) c E 0 ( x ^ − y ^ ) cos ( k z − ω t ) (d) E 0 c ( − x ^ + y ^ ) sin ( k z − ω t ) \frac{E_0}{c}\left(-\hat{x} + \hat{y}\right)\sin(kz - \omega t) c E 0 ( − x ^ + y ^ ) sin ( k z − ω t ) Show answer and solution
Answer: (d)
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Q53 JEE Main 2020 · 3 SEP · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is E ⃗ = E 0 j ^ cos ( ω t − k x ) \vec{E} = E_0\hat{j}\cos(\omega t - kx) E = E 0 j ^ cos ( ω t − k x ) .
The magnetic field B ⃗ \vec{B} B , at the moment t = 0 is:
(a) B ⃗ = E 0 μ 0 ε 0 cos ( k x ) j ^ \vec{B} = \frac{E_0}{\sqrt{\mu_0\varepsilon_0}}\cos(kx)\hat{j} B = μ 0 ε 0 E 0 cos ( k x ) j ^ (b) B ⃗ = E 0 μ 0 ε 0 cos ( k x ) k ^ \vec{B} = \frac{E_0}{\sqrt{\mu_0\varepsilon_0}}\cos(kx)\hat{k} B = μ 0 ε 0 E 0 cos ( k x ) k ^ (c) B ⃗ = E 0 μ 0 ε 0 cos ( k x ) k ^ \vec{B} = E_0\sqrt{\mu_0\varepsilon_0}\cos(kx)\hat{k} B = E 0 μ 0 ε 0 cos ( k x ) k ^ (d) B ⃗ = E 0 μ 0 ε 0 cos ( k x ) j ^ \vec{B} = E_0\sqrt{\mu_0\varepsilon_0}\cos(kx)\hat{j} B = E 0 μ 0 ε 0 cos ( k x ) j ^ Show answer and solution
Answer: (c)
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Q54 JEE Main 2020 · 3 SEP · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The magnetic field of a plane electromagnetic wave is
B ⃗ = 3 × 1 0 − 8 sin [ 200 π ( y + c t ) ] i ^ \vec{B} = 3\times 10^{-8}\sin\left[200\pi(y + ct)\right]\hat{i} B = 3 × 1 0 − 8 sin [ 200 π ( y + c t ) ] i ^ T
where c = 3 × \times × 108 ms–1 is the speed of light. The corresponding electric field is:
(a) E ⃗ = − 1 0 − 6 sin [ 200 π ( y + c t ) ] k ^ \vec{E} = -10^{-6}\sin\left[200\pi(y + ct)\right]\hat{k} E = − 1 0 − 6 sin [ 200 π ( y + c t ) ] k ^ V/m(b) E ⃗ = − 9 sin [ 200 π ( y + c t ) ] k ^ \vec{E} = -9\sin\left[200\pi(y + ct)\right]\hat{k} E = − 9 sin [ 200 π ( y + c t ) ] k ^ V/m(c) E ⃗ = 9 sin [ 200 π ( y + c t ) ] k ^ \vec{E} = 9\sin\left[200\pi(y + ct)\right]\hat{k} E = 9 sin [ 200 π ( y + c t ) ] k ^ V/m(d) E ⃗ = 3 × 1 0 − 8 sin [ 200 π ( y + c t ) ] k ^ \vec{E} = 3\times 10^{-8}\sin\left[200\pi(y + ct)\right]\hat{k} E = 3 × 1 0 − 8 sin [ 200 π ( y + c t ) ] k ^ Show answer and solution
Answer: (b)
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Q55 JEE Main 2020 · 8 JAN · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by B ⃗ = 5 × 1 0 − 8 j ^ T \vec{B} = 5\times 10^{-8}\hat{j}\text{ T} B = 5 × 1 0 − 8 j ^ T . The corresponding electric field E ⃗ \vec{E} E is (speed of light c = 3 × 108 ms–1)
(a) 15 i ^ \hat{i} i ^ V / m (b) -15 i ^ \hat{i} i ^ V / m (c) 1.66 × 10–16 i ^ \hat{i} i ^ V / m (d) -1.66 × 10–16 i ^ \hat{i} i ^ V / m Show answer and solution
Answer: (a)
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Q56 JEE Main 2020 · 7 JAN · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
If the magnetic field in a plane electromagnetic wave is given by
B ⃗ \vec{B} B = 3 × \times × 10-8 sin(1.6 × \times × 103x + 48 × \times × 1010t)j ^ \hat{j} j ^ T, then what will be expression for electric field?
(a) E ⃗ \vec{E} E = (9sin(1.6 × \times × 103x + 48 × \times × 1010t)k ^ \hat{k} k ^ V/m)(b) E ⃗ \vec{E} E = (60sin(1.6 × \times × 103x + 48 × \times × 1010t)k ^ \hat{k} k ^ V/m)(c) E ⃗ \vec{E} E = (3 × \times × 10-8 sin(1.6 × \times × 103x + 48 × \times × 1010t)i ^ \hat{i} i ^ V/m)(d) E ⃗ \vec{E} E = (3 × \times × 10-8 sin(1.6 × \times × 103x + 48 × \times × 1010t)j ^ \hat{j} j ^ V/m)Show answer and solution
Answer: (a)
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Q57 JEE Main 2020 · 5 SEP · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Force on a charge
An electron is constrained to move along the y-axis with a speed of 0.1 c (c is the speed of light) in the presence of electromagnetic wave, whose electric field is
E ⃗ = 30 j ^ sin ( 1.5 × 1 0 7 t − 5 × 1 0 − 2 x ) \vec{E} = 30\hat{j} \sin\left(1.5 \times 10^{7} t - 5 \times 10^{-2} x\right) E = 30 j ^ sin ( 1.5 × 1 0 7 t − 5 × 1 0 − 2 x ) V/m.
The maximum magnetic force experienced by the electron will be:
(given c = 3 × \times × 108 ms–1 and electron charge = 1.6 × \times × 10–19 C)
(a) 4.8 × \times × 10–19 N (b) 2.4 × \times × 10–18 N (c) 3.2 × \times × 10–18 N (d) 1.6 × \times × 10–18 N Show answer and solution
Answer: (a)
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Q58 JEE Main 2020 · 9 JAN · SHIFT 2 Difficult ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
A plane electromagnetic wave is propagating along the direction i ^ + j ^ 2 \frac{\hat{i} + \hat{j}}{\sqrt{2}} 2 i ^ + j ^ , with its polarization along the direction k ^ \hat{k} k ^ . The correct form of the magnetic field of the wave would be (here B0 is an appropriate constant):
(a) B 0 i ^ − j ^ 2 cos ( ω t − k i ^ + j ^ 2 ) B_0 \frac{\hat{i} - \hat{j}}{\sqrt{2}} \cos\left(\omega t - k\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) B 0 2 i ^ − j ^ cos ( ω t − k 2 i ^ + j ^ ) (b) B 0 i ^ + j ^ 2 cos ( ω t − k i ^ + j ^ 2 ) B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos\left(\omega t - k\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) B 0 2 i ^ + j ^ cos ( ω t − k 2 i ^ + j ^ ) (c) B 0 j ^ − i ^ 2 cos ( ω t + k i ^ + j ^ 2 ) B_0 \frac{\hat{j} - \hat{i}}{\sqrt{2}} \cos\left(\omega t + k\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) B 0 2 j ^ − i ^ cos ( ω t + k 2 i ^ + j ^ ) (d) B 0 k ^ cos ( ω t − k i ^ + j ^ 2 ) B_0 \hat{k} \cos\left(\omega t - k\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) B 0 k ^ cos ( ω t − k 2 i ^ + j ^ ) Show answer and solution
Answer: (a)
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Q59 JEE Main 2020 · 9 JAN · SHIFT 1 Difficult ☆
Nature of EM Waves & Wave Equation · Force on a charge
The electric fields of two plane electromagnetic plane waves in vacuum are given by and
At t = 0, a particle of charge q is at origin with
a velocityv ⃗ = 0.8 c j ^ \vec{v} = 0.8c\hat{j} v = 0.8 c j ^ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is:
(a) E 0 q ( 0.8 i ^ − j ^ + 0.4 k ^ ) E_0 q \left(0.8\hat{i} - \hat{j} + 0.4\hat{k}\right) E 0 q ( 0.8 i ^ − j ^ + 0.4 k ^ ) (b) E 0 q ( − 0.8 i ^ + j ^ + k ^ ) E_0 q \left(-0.8\hat{i} + \hat{j} + \hat{k}\right) E 0 q ( − 0.8 i ^ + j ^ + k ^ ) (c) E 0 q ( 0.8 i ^ + j ^ + 0.2 k ^ ) E_0 q \left(0.8\hat{i} + \hat{j} + 0.2\hat{k}\right) E 0 q ( 0.8 i ^ + j ^ + 0.2 k ^ ) (d) E 0 q ( 0.4 i ^ − 3 j ^ + 0.8 k ^ ) E_0 q \left(0.4\hat{i} - 3\hat{j} + 0.8\hat{k}\right) E 0 q ( 0.4 i ^ − 3 j ^ + 0.8 k ^ ) Show answer and solution
Answer: (c)
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Q60 JEE Main 2020 · 7 JAN · SHIFT 2 Difficult ☆
Nature of EM Waves & Wave Equation · Force on a charge
The electric field of a plane electromagnetic wave is given by
E ⃗ = E 0 i ^ + j ^ 2 cos ( k z + ω t ) \vec{E} = E_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz + \omega t) E = E 0 2 i ^ + j ^ cos ( k z + ω t )
At t = 0, a positively charged particle is at the point (x, y, z) = ( 0 , 0 , π k ) \left(0, 0, \frac{\pi}{k}\right) ( 0 , 0 , k π ) .
If its instantaneous velocity at (t = 0) is v 0 k ^ v_0 \hat{k} v 0 k ^ , the force acting on it due to the wave is:
(a) parallel to k ^ \hat{k} k ^ (b) parallel to i ^ + j ^ 2 \frac{\hat{i} + \hat{j}}{\sqrt{2}} 2 i ^ + j ^ (c) antiparallel to i ^ + j ^ 2 \frac{\hat{i} + \hat{j}}{\sqrt{2}} 2 i ^ + j ^ (d) zero Show answer and solution
Answer: (c)
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Q61 JEE Main 2019 · 12 APR · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
An electromagnetic wave is represented by the electric field E ⃗ = E 0 n ^ sin [ ω t + ( 6 y − 8 z ) ] \vec{E} = E_0\hat{n}\sin\left[\omega t + (6y - 8z)\right] E = E 0 n ^ sin [ ω t + ( 6 y − 8 z ) ] . Taking unit vectors in x, y and z directions to be i ^ , j ^ , k ^ \hat{i}, \hat{j}, \hat{k} i ^ , j ^ , k ^ , the direction of propagation s ^ \hat{s} s ^ , is:
(a) s ^ = 3 i ^ − 4 j ^ 5 \hat{s} = \frac{3\hat{i}-4\hat{j}}{5} s ^ = 5 3 i ^ − 4 j ^ (b) s ^ = − 4 k ^ + 3 j ^ 5 \hat{s} = \frac{-4\hat{k}+3\hat{j}}{5} s ^ = 5 − 4 k ^ + 3 j ^ (c) s ^ = ( − 3 j ^ + 4 k ^ 5 ) \hat{s} = \left(\frac{-3\hat{j}+4\hat{k}}{5}\right) s ^ = ( 5 − 3 j ^ + 4 k ^ ) (d) s ^ = 4 j ^ − 3 k ^ 5 \hat{s} = \frac{4\hat{j}-3\hat{k}}{5} s ^ = 5 4 j ^ − 3 k ^ Show answer and solution
Answer: (c)
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Q62 JEE Main 2019 · 8 APR · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is E = 6 V m–1 along y-direction. Its corresponding magnetic field component, B would be:
(a) 2 × 10–8 T along y-direction (b) 6 × 10–8 T along z-direction (c) 2 × 10–8 T along z-direction (d) 6 × 10–8 T along x-direction Show answer and solution
Answer: (c)
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Q63 JEE Main 2019 · 10 JAN · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 m/s) B = 100 × 10–6 sin [ 2 π × 2 × 1 0 15 ( t − x c ) ] \left[2\pi\times 2\times 10^{15}\left(t - \frac{x}{c}\right)\right] [ 2 π × 2 × 1 0 15 ( t − c x ) ] then the maximum electric field associated with it is -
(a) 4.5 × \times × 104 N/C (b) 4 × \times × 104 N/C (c) 6 × \times × 104 N/C (d) 3 × \times × 104 N/C Show answer and solution
Answer: (d)
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Q64 JEE Main 2019 · 9 JAN · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x- direction. At a particular point in space and time, E ⃗ = 6.3 j ^ V/m \vec{E} = 6.3\hat{j}\text{ V/m} E = 6.3 j ^ V/m The corresponding magnetic field B ⃗ \vec{B} B at that point will be:
(a) 18.9 × \times × 10− - − 8 k ^ \hat{k} k ^ T (b) 2.1 × \times × 10− - − 8 k ^ \hat{k} k ^ T (c) 6.3 × \times × 10− - − 8 k ^ \hat{k} k ^ T (d) 18.9 × \times × 108 k ^ \hat{k} k ^ T Show answer and solution
Answer: (b)
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Q65 JEE Main 2019 · 12 APR · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
A plane electromagnetic wave having a frequency v = 23.9 GHz propagates along the positive z- direction in free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave?
(a) B ⃗ \vec{B} B = 2 × 10–7 sin(1.5 × 102 x + 0.5 × 1011t) j ^ \hat{j} j ^ (b) B ⃗ \vec{B} B = 60 sin(0.5 × 103x + 0.5 × 1011t) k ^ \hat{k} k ^ (c) B ⃗ \vec{B} B = 2 × 10–7 sin(0.5 × 103 z + 1.5 × 1011t) i ^ \hat{i} i ^ (d) B ⃗ \vec{B} B = 2 × 10–7 sin(0.5 × 103 z - 1.5 × 1011t) i ^ \hat{i} i ^ Show answer and solution
Answer: (d)
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Q66 JEE Main 2019 · 10 APR · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field of a plane electromagnetic wave is given by
E ⃗ = E 0 i ^ cos ( k z ) cos ( ω t ) \vec{E} = E_0 \hat{i} \cos(kz) \cos(\omega t) E = E 0 i ^ cos ( k z ) cos ( ω t )
The corresponding magnetic field B ⃗ \vec{B} B is then given by
(a) B ⃗ = E 0 C j ^ sin ( k z ) sin ( ω t ) \vec{B} = \frac{E_0}{C} \hat{j} \sin(kz) \sin(\omega t) B = C E 0 j ^ sin ( k z ) sin ( ω t ) (b) B ⃗ = E 0 C j ^ sin ( k z ) cos ( ω t ) \vec{B} = \frac{E_0}{C} \hat{j} \sin(kz) \cos(\omega t) B = C E 0 j ^ sin ( k z ) cos ( ω t ) (c) B ⃗ = E 0 C j ^ cos ( k z ) sin ( ω t ) \vec{B} = \frac{E_0}{C} \hat{j} \cos(kz) \sin(\omega t) B = C E 0 j ^ cos ( k z ) sin ( ω t ) (d) B ⃗ = E 0 C k ^ sin ( k z ) cos ( ω t ) \vec{B} = \frac{E_0}{C} \hat{k} \sin(kz) \cos(\omega t) B = C E 0 k ^ sin ( k z ) cos ( ω t ) Show answer and solution
Answer: (a)
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Q67 JEE Main 2019 · 8 APR · SHIFT 2 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The magnetic field of an electromagnetic wave is given by :-
B ⃗ = 1.6 × 1 0 − 6 cos ( 2 × 1 0 7 z + 6 × 1 0 15 t ) ( 2 i ^ + j ^ ) Wb m 2 \vec{B} = 1.6 \times 10^{-6} \cos\left(2 \times 10^{7} z + 6 \times 10^{15} t\right) \left(2\hat{i} + \hat{j}\right) \frac{\text{Wb}}{\text{m}^2} B = 1.6 × 1 0 − 6 cos ( 2 × 1 0 7 z + 6 × 1 0 15 t ) ( 2 i ^ + j ^ ) m 2 Wb
The associated electric field will be :-
(a) E ⃗ = 4.8 × 1 0 2 cos ( 2 × 1 0 7 z − 6 × 1 0 15 t ) ( − 2 i ^ + j ^ ) V m \vec{E} = 4.8 \times 10^{2} \cos\left(2 \times 10^{7} z - 6 \times 10^{15} t\right) \left(-2\hat{i} + \hat{j}\right) \frac{\text{V}}{\text{m}} E = 4.8 × 1 0 2 cos ( 2 × 1 0 7 z − 6 × 1 0 15 t ) ( − 2 i ^ + j ^ ) m V (b) E ⃗ = 4.8 × 1 0 2 cos ( 2 × 1 0 7 z − 6 × 1 0 15 t ) ( 2 i ^ + j ^ ) V m \vec{E} = 4.8 \times 10^{2} \cos\left(2 \times 10^{7} z - 6 \times 10^{15} t\right) \left(2\hat{i} + \hat{j}\right) \frac{\text{V}}{\text{m}} E = 4.8 × 1 0 2 cos ( 2 × 1 0 7 z − 6 × 1 0 15 t ) ( 2 i ^ + j ^ ) m V (c) E ⃗ = 4.8 × 1 0 2 cos ( 2 × 1 0 7 z + 6 × 1 0 15 t ) ( i ^ − 2 j ^ ) V m \vec{E} = 4.8 \times 10^{2} \cos\left(2 \times 10^{7} z + 6 \times 10^{15} t\right) \left(\hat{i} - 2\hat{j}\right) \frac{\text{V}}{\text{m}} E = 4.8 × 1 0 2 cos ( 2 × 1 0 7 z + 6 × 1 0 15 t ) ( i ^ − 2 j ^ ) m V (d) E ⃗ = 4.8 × 1 0 2 cos ( 2 × 1 0 7 z + 6 × 1 0 15 t ) ( − i ^ + 2 j ^ ) V m \vec{E} = 4.8 \times 10^{2} \cos\left(2 \times 10^{7} z + 6 \times 10^{15} t\right) \left(-\hat{i} + 2\hat{j}\right) \frac{\text{V}}{\text{m}} E = 4.8 × 1 0 2 cos ( 2 × 1 0 7 z + 6 × 1 0 15 t ) ( − i ^ + 2 j ^ ) m V Show answer and solution
Answer: (c)
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Q68 JEE Main 2019 · 10 JAN · SHIFT 2 Difficult ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression E ⃗ ( x , y ) = 10 j ^ cos [ ( 6 x + 8 z ) ] . \vec{E}(x, y) = 10\hat{j} \cos[(6x + 8z)]. E ( x , y ) = 10 j ^ cos [( 6 x + 8 z )] . The magnetic field B ⃗ \vec{B} B (x,z, t) is given by − - − (c is the velocity of light)
(a) 1 c ( 6 k ^ + 8 i ^ ) cos [ ( 6 x + 8 z − 10 c t ) ] \frac{1}{c}\left(6\hat{k} + 8\hat{i}\right) \cos[(6x + 8z - 10ct)] c 1 ( 6 k ^ + 8 i ^ ) cos [( 6 x + 8 z − 10 c t )] (b) 1 c ( 6 k ^ − 8 i ^ ) cos [ ( 6 x + 8 z − 10 c t ) ] \frac{1}{c}\left(6\hat{k} - 8\hat{i}\right) \cos[(6x + 8z - 10ct)] c 1 ( 6 k ^ − 8 i ^ ) cos [( 6 x + 8 z − 10 c t )] (c) 1 c ( 6 k ^ + 8 i ^ ) cos [ ( 6 x − 8 z + 10 c t ) ] \frac{1}{c}\left(6\hat{k} + 8\hat{i}\right) \cos[(6x - 8z + 10ct)] c 1 ( 6 k ^ + 8 i ^ ) cos [( 6 x − 8 z + 10 c t )] (d) 1 c ( 6 k ^ − 8 i ^ ) cos [ ( 6 x − 8 z + 10 c t ) ] \frac{1}{c}\left(6\hat{k} - 8\hat{i}\right) \cos[(6x - 8z + 10ct)] c 1 ( 6 k ^ − 8 i ^ ) cos [( 6 x − 8 z + 10 c t )] Show answer and solution
Answer: (b)
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Q69 JEE Main 2019 · 9 APR · SHIFT 1 Difficult ☆
Nature of EM Waves & Wave Equation · Force on a charge
The magnetic field of a plane electromagnetic wave is given by:
B ⃗ = B 0 i ^ [ cos ( k z − ω t ) ] + B 0 j ^ cos ( k z + ω t ) \vec{B} = B_0 \hat{i} [\cos(kz - \omega t)] + B_0 \hat{j} \cos(kz + \omega t) B = B 0 i ^ [ cos ( k z − ω t )] + B 0 j ^ cos ( k z + ω t )
B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to:
(a) 0.6 N (b) 0.9 N (c) 3 × 10–2 N (d) 0.1 N Show answer and solution
Answer: (a)
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Q70 NEET 2018 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
An EM wave is propagating in a medium with a velocityv ⃗ = v i ^ \vec{v} = v\hat{i} v = v i ^ . The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along
(a) –z direction (b) +z direction (c) –y direction (d) –x direction Show answer and solution
Answer: (b)
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Q71 JEE Main 2018 · 16 APR · SHIFT 1 Difficult ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
A plane electromagnetic wave of wavelength λ \lambda λ has an intensity I. It is propagatting along the positive Y-direction. The allowed expressions for the electric and magnetic fields are givn by:
Show answer and solution
Answer: (a)
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Q72 JEE Main 2018 · 15 APR · SHIFT 1 Difficult ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
A monochromatic beam of light has a frequency ν = 3 2 π × 1 0 12 Hz \nu = \frac{3}{2\pi} \times 10^{12}\text{ Hz} ν = 2 π 3 × 1 0 12 Hz and is propagating along the direction i ^ + j ^ 2 . \frac{\hat{i} + \hat{j}}{\sqrt{2}}. 2 i ^ + j ^ .
It is polarized along the k ^ \hat{k} k ^ direction. The acceptable form for the magnetic field is:
Show answer and solution
Answer: (a)
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Q73 JEE Main 2018 · 15 APR · SHIFT 2 Difficult ☆
Nature of EM Waves & Wave Equation · Reading the wave equation
A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electromagnetic wave is:
(a) 3 × 1 0 8 ∣ z 2 − z 1 ∣ \frac{3 \times 10^{8}}{|z_2 - z_1|} ∣ z 2 − z 1 ∣ 3 × 1 0 8 (b) 1.5 × 1 0 8 ∣ z 2 − z 1 ∣ \frac{1.5 \times 10^{8}}{|z_2 - z_1|} ∣ z 2 − z 1 ∣ 1.5 × 1 0 8 (c) 6 × 1 0 8 ∣ z 2 − z 1 ∣ \frac{6 \times 10^{8}}{|z_2 - z_1|} ∣ z 2 − z 1 ∣ 6 × 1 0 8 (d) 1 t 1 + ∣ z 2 − z 1 ∣ 3 × 1 0 8 \frac{1}{t_1 + \frac{|z_2 - z_1|}{3 \times 10^{8}}} t 1 + 3 × 1 0 8 ∣ z 2 − z 1 ∣ 1 Show answer and solution
Answer: (b)
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Q74 NEET 2017 Medium ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
In an electromagnetic wave in free space that root mean square value of the electric field is Erms = 6 V m− 1 {}^{-1} − 1 . The peak value of the magnetic field is
(a) 2.83 × \times × 10− - − 8 T (b) 0.70 × \times × 10− - − 8 T (c) 4.23 × \times × 10− - − 8 T (d) 1.41 × \times × 10− - − 8 T Show answer and solution
Answer: (a)
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Q75 JEE Main 2017 · 8 APR · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
Magnetic field in a plane electromagnetic wave is given by
B ⃗ \vec{B} B = B0 sin (k x + ω \omega ω t) j ^ T \hat{j}\,\text{T} j ^ T
Expression for corresponding electric field will be:
Where c is speed of light.
(a) E ⃗ \vec{E} E = B0 c sin (k x + ω \omega ω t) k ^ \hat{k} k ^ V/m(b) E ⃗ = B 0 c \vec{E} = \frac{B_0}{c} E = c B 0 sin (k x + ω \omega ω t) k ^ \hat{k} k ^ V/m(c) E ⃗ = − \vec{E} = - E = − B0 c sin (kx +ω \omega ω t) k ^ \hat{k} k ^ V/m(d) E ⃗ \vec{E} E = B0 c sin (kx − ω -\omega − ω t) k ^ \hat{k} k ^ V/mShow answer and solution
Answer: (a)
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Q76 JEE Main 2017 · 9 APR · SHIFT 1 Difficult ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
The electric field component of a monochromatic radiation is given by
E ⃗ \vec{E} E = 2 E0 i ^ \hat{i} i ^ cos kz cos ω \omega ω t
Its magnetic field B ⃗ \vec{B} B is then given by:
(a) 2 E 0 c j ^ \frac{2E_0}{c} \hat{j} c 2 E 0 j ^ sin kz cos ω \omega ω t(b) − 2 E 0 c j ^ -\frac{2E_0}{c} \hat{j} − c 2 E 0 j ^ sin kz sin ω \omega ω t(c) 2 E 0 c j ^ \frac{2E_0}{c} \hat{j} c 2 E 0 j ^ sin kz sin ω \omega ω t(d) 2 E 0 c j ^ \frac{2E_0}{c} \hat{j} c 2 E 0 j ^ cos kz cos ω \omega ω tShow answer and solution
Answer: (c)
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Q77 JEE Main 2016 · 10 APR · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
Consider an electromagnetic wave propagating in vacuum. Choose the correct statement:
(a) For an electromagnetic wave propagating in +x direction the electric field is E ⃗ = 1 2 E y z ( x , t ) ( y ^ − z ^ ) \vec{E} = \frac{1}{\sqrt{2}}E_{yz}(x,t)\left(\hat{y} - \hat{z}\right) E = 2 1 E yz ( x , t ) ( y ^ − z ^ ) and the magnetic field is B ⃗ = 1 2 B y z ( x , t ) ( y ^ + z ^ ) \vec{B} = \frac{1}{\sqrt{2}}B_{yz}(x,t)\left(\hat{y} + \hat{z}\right) B = 2 1 B yz ( x , t ) ( y ^ + z ^ ) (b) For an electromagnetic wave propagating in +x direction the electric field is E ⃗ = 1 2 E y z ( y , z , t ) ( y ^ + z ^ ) \vec{E} = \frac{1}{\sqrt{2}}E_{yz}(y,z,t)\left(\hat{y} + \hat{z}\right) E = 2 1 E yz ( y , z , t ) ( y ^ + z ^ ) and the magnetic field is B ⃗ = 1 2 B y z ( y , z , t ) ( y ^ + z ^ ) \vec{B} = \frac{1}{\sqrt{2}}B_{yz}(y,z,t)\left(\hat{y} + \hat{z}\right) B = 2 1 B yz ( y , z , t ) ( y ^ + z ^ ) (c) For an electromagnetic wave propagating in + y direction the electric field is E ⃗ = 1 2 E y z ( x , t ) y ^ \vec{E} = \frac{1}{\sqrt{2}}E_{yz}(x,t)\hat{y} E = 2 1 E yz ( x , t ) y ^ and the magnetic field is B ⃗ = 1 2 B y z ( x , t ) z ^ \vec{B} = \frac{1}{\sqrt{2}}B_{yz}(x,t)\hat{z} B = 2 1 B yz ( x , t ) z ^ (d) For an electromagnetic wave propagating in + y direction the electric field is E ⃗ = 1 2 E y z ( x , t ) z ^ \vec{E} = \frac{1}{\sqrt{2}}E_{yz}(x,t)\hat{z} E = 2 1 E yz ( x , t ) z ^ and the magnetic field is B ⃗ = 1 2 B z ( x , t ) y ^ \vec{B} = \frac{1}{\sqrt{2}}B_{z}(x,t)\hat{y} B = 2 1 B z ( x , t ) y ^ Show answer and solution
Answer: (a)
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Q78 JEE Main 2015 · 11 APR · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
For plane electromagnetic waves propagating in the z direction, which one of the following combination gives the correct possible direction for E ⃗ \vec{E} E and B ⃗ \vec{B} B field respectively?
(a) ( i ^ + 2 j ^ ) \left(\hat{i}+2\hat{j}\right) ( i ^ + 2 j ^ ) and ( 2 i ^ − j ^ ) \left(2\hat{i}-\hat{j}\right) ( 2 i ^ − j ^ ) (b) ( − 2 i ^ − 3 j ^ ) \left(-2\hat{i}-3\hat{j}\right) ( − 2 i ^ − 3 j ^ ) and ( 3 i ^ − 2 j ^ ) \left(3\hat{i}-2\hat{j}\right) ( 3 i ^ − 2 j ^ ) (c) ( 2 i ^ + 3 j ^ ) \left(2\hat{i}+3\hat{j}\right) ( 2 i ^ + 3 j ^ ) and ( i ^ + 2 j ^ ) \left(\hat{i}+2\hat{j}\right) ( i ^ + 2 j ^ ) (d) ( 3 i ^ + 4 j ^ ) \left(3\hat{i}+4\hat{j}\right) ( 3 i ^ + 4 j ^ ) and ( 4 i ^ − 3 j ^ ) \left(4\hat{i}-3\hat{j}\right) ( 4 i ^ − 3 j ^ ) Show answer and solution
Answer: (b)
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Q79 JEE Main 2015 · 10 APR · SHIFT 1 Medium ☆
Nature of EM Waves & Wave Equation · Writing B from E and E from B
An electromagnetic wave travelling in the x-direction has frequency of 2 × \times × 1014 Hz and electric field amplitude of 27 Vm− 1 {}^{-1} − 1 . From the options given below, which one describes the magnetic field for this wave?
(a) B ⃗ \vec{B} B (x, t) = (3 × \times × 10− - − 8T) j ^ \hat{j} j ^ sin [ 2π \pi π (1.5 × \times × 10 − - − 8x − - − 2 × \times × 1014t)](b) B ⃗ \vec{B} B (x, t) = (9 × \times × 10− - − 8T) k ^ \hat{k} k ^ sin [ 2π \pi π (1.5 × \times × 10− - − 6x − - − 2 × \times × 1014t)](c) B ⃗ \vec{B} B (x, t) = (9 × \times × 10− - − 8T) i ^ \hat{i} i ^ sin [ 2π \pi π (1.5 × \times × 10− - − 8x − - − 2 × \times × 1014t)](d) B ⃗ \vec{B} B (x, t) = (9 × \times × 10− - − 8T) j ^ \hat{j} j ^ sin [(1.5 × \times × 10− - − 6 x − - − 2 × \times × 1014t)]Show answer and solution
Answer: (b)
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Q80 JEE Main 2013 · 23 APR · SHIFT 1 Easy ☆
Nature of EM Waves & Wave Equation · Nature of EM waves
Select the correct statement from the following:
(a) Electromagnetic waves cannot travel in vacuum. (b) Electromagnetic waves are longitudinal waves. (c) Electromagnetic waves are produced by charges moving with uniform velocity. (d) Electromagnetic waves carry both energy and momentum as they propagate through space. Show answer and solution
Answer: (d)
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Q81 JEE Main 2013 Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT 20\text{ nT} 20 nT . The peak value of electric field strength is:
(a) 3 V/m 3\text{ V/m} 3 V/m (b) 6 V/m 6\text{ V/m} 6 V/m (c) 9 V/m 9\text{ V/m} 9 V/m (d) 12 V/m 12\text{ V/m} 12 V/m Show answer and solution
Answer: (b)
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Q82 JEE Main 2012 · AIEEE Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
An electromagnetic wave in vacuum has the electric and magnetic field E ⃗ \vec{E} E and B ⃗ \vec{B} B , which are always perpendicular to each other. The direction of polarization is given by X ⃗ \vec{X} X and that of wave propagation byk ⃗ \vec{k} k . Then
(a) X ⃗ ∥ B ⃗ \vec{X}\parallel\vec{B} X ∥ B and X ⃗ ∥ B ⃗ × E ⃗ \vec{X}\parallel\vec{B}\times\vec{E} X ∥ B × E (b) X ⃗ ∥ E ⃗ \vec{X}\parallel\vec{E} X ∥ E andk ⃗ ∥ E ⃗ × B ⃗ \vec{k}\parallel\vec{E}\times\vec{B} k ∥ E × B (c) X ⃗ ∥ B ⃗ \vec{X}\parallel\vec{B} X ∥ B andk ⃗ ∥ E ⃗ × B ⃗ \vec{k}\parallel\vec{E}\times\vec{B} k ∥ E × B (d) X ⃗ ∥ E ⃗ \vec{X}\parallel\vec{E} X ∥ E andk ⃗ ∥ B ⃗ × E ⃗ \vec{k}\parallel\vec{B}\times\vec{E} k ∥ B × E Show answer and solution
Answer: (b)
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Q83 NEET 2012 · AIPMT · PRELIMS Easy ☆
Nature of EM Waves & Wave Equation · Reading the wave equation
The electric field associated with an em wave in vacuum is given by
E = i ^ \hat{i} i ^ 40 cos (kz − - − 6 × \times × 108 t) where E, z and t are in volt/m, meter and seconds respectively. The value of wave vector k is
(a) 2 m− 1 {}^{-1} − 1 (b) 0.5 m− 1 {}^{-1} − 1 (c) 6 m− 1 {}^{-1} − 1 (d) 3 m− 1 {}^{-1} − 1 Show answer and solution
Answer: (a)
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Q84 NEET 2012 · AIPMT · MAINS Easy ☆
Nature of EM Waves & Wave Equation · Field values: E0 = cB0
The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagatting in vacuum is equal to
(a) the speed of light in vacuum (b) reciprocal of speed of light in vacuum (c) the ratio of magnetic permeability to the electric susceptibility of vacuum (d) unity Show answer and solution
Answer: (b)
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Q85 NEET 2011 · AIPMT · PRELIMS Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
The electric and the magnetic field, associated with an e.m. wave, propagating along the +z-axis, can be represented by
(a) [ E ⃗ = E 0 i ^ , B ⃗ = B 0 j ^ ] \left[\vec{E} = E_0\hat{i},\ \vec{B} = B_0\hat{j}\right] [ E = E 0 i ^ , B = B 0 j ^ ] (b) [ E ⃗ = E 0 k ^ , B ⃗ = B 0 i ^ ] \left[\vec{E} = E_0\hat{k},\ \vec{B} = B_0\hat{i}\right] [ E = E 0 k ^ , B = B 0 i ^ ] (c) [ E ⃗ = E 0 j ^ , B ⃗ = B 0 i ^ ] \left[\vec{E} = E_0\hat{j},\ \vec{B} = B_0\hat{i}\right] [ E = E 0 j ^ , B = B 0 i ^ ] (d) [ E ⃗ = E 0 j ^ , B ⃗ = B 0 k ^ ] \left[\vec{E} = E_0\hat{j},\ \vec{B} = B_0\hat{k}\right] [ E = E 0 j ^ , B = B 0 k ^ ] Show answer and solution
Answer: (a)
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Q86 NEET 2010 · AIPMT · PRELIMS Easy ☆
Nature of EM Waves & Wave Equation · Nature of EM waves
Which of the following statements is false for the properties of electromagnetic waves?
(a) Both electric and magnetic field vectors attain the maxima and minima at the same place and same time. (b) The energy in electromagnetic wave is divided equally between electric and magnetic vectors. (c) Both electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation of wave. (d) These waves do not require any material medium for propagation. Show answer and solution
Answer: (c)
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Q87 NEET 2010 · AIPMT · MAINS Medium ☆
Nature of EM Waves & Wave Equation · Reading the wave equation
The electric field of an electromagnetic wave in free space is given by E ⃗ = 10 cos ( 1 0 7 t + k x ) j ^ V/m , \vec{E} = 10\cos(10^{7}t + kx)\hat{j}\text{ V/m}, E = 10 cos ( 1 0 7 t + k x ) j ^ V/m , where t and x are in seconds and metres respectively. It can be inferred that
(1) the wavelength λ \lambda λ is 188.4 m.
(2) the wave number k is 0.33 rad/m.
(3) the wave amplitude is 10 V/m.
(4) the wave is propagating along +x direction.
Which one of the following pairs of statements is correct?
(a) (3) and (4) (b) (1) and (2) (c) (2) and (3) (d) (1) and (3) Show answer and solution
Answer: (d)
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Q88 NEET 2009 · AIPMT Medium ☆
Nature of EM Waves & Wave Equation · Reading the wave equation
The electric field part of an electromagnetic wave in a medium is represented by Ex = 0; E y = 2.5 N C cos [ ( 2 π × 1 0 6 rad s ) t − ( π × 1 0 − 2 rad m ) x ] ; E_y = 2.5\frac{\text{N}}{\text{C}}\cos\left[\left(2\pi\times 10^{6}\frac{\text{rad}}{\text{s}}\right)t - \left(\pi\times 10^{-2}\frac{\text{rad}}{\text{m}}\right)x\right]; E y = 2.5 C N cos [ ( 2 π × 1 0 6 s rad ) t − ( π × 1 0 − 2 m rad ) x ] ;
Ez = 0. The wave is
(a) moving along x direction with frequency 106 Hz and wavelength 100 m. (b) moving along x direction with frequency 106 Hz and wavelength 200 m. (c) moving along − - − x direction with frequency 106 Hz and wavelength 200 m. (d) moving along y direction with frequency 2 π × 1 0 6 Hz 2\pi\times 10^{6}\text{ Hz} 2 π × 1 0 6 Hz and wavelength 200 m. Show answer and solution
Answer: (b)
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Q89 NEET 2007 · AIPMT Easy ☆
Nature of EM Waves & Wave Equation · Nature of EM waves
The electric and magnetic field of an electromagnetic wave are
(a) in opposite phase and perpendicular to each other (b) in opposite phase and parallel to each other (c) in phase and perpendicular to each other (d) in phase and parallel to each other. Show answer and solution
Answer: (c)
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Q90 JEE Main 2002 · AIEEE Easy ☆
Nature of EM Waves & Wave Equation · Nature of EM waves
Electromagnetic waves are transverse in nature is evident by
(a) polarization (b) interference (c) reflection (d) diffraction Show answer and solution
Answer: (a)
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Q91 NEET 2002 · AIPMT Easy ☆
Nature of EM Waves & Wave Equation · Direction of E, B and propagation
The velocity of electromagnetic wave is parallel to
(a) B ⃗ × E ⃗ \vec{B}\times\vec{E} B × E (b) E ⃗ × B ⃗ \vec{E}\times\vec{B} E × B (c) E ⃗ \vec{E} E (d) B ⃗ \vec{B} B Show answer and solution
Answer: (b)
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Q92 NEET 1999 · AIPMT Easy ☆
Nature of EM Waves & Wave Equation · Reading the wave equation
The wavelength of light of frequency 100 Hz 100\text{ Hz} 100 Hz is
(a) 2 × 1 0 6 m 2\times 10^{6}\text{ m} 2 × 1 0 6 m (b) 3 × 1 0 6 m 3\times 10^{6}\text{ m} 3 × 1 0 6 m (c) 4 × 1 0 6 m 4\times 10^{6}\text{ m} 4 × 1 0 6 m (d) 5 × 1 0 6 m 5\times 10^{6}\text{ m} 5 × 1 0 6 m Show answer and solution
Answer: (b)
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Q93 NEET 1994 · AIPMT Easy ☆
Nature of EM Waves & Wave Equation · Nature of EM waves
The electric and magnetic field of an electromagnetic wave are
(a) in opposite phase and perpendicular to each other (b) in opposite phase and parallel to each other (c) in phase and perpendicular to each other (d) in phase and parallel to each other Show answer and solution
Answer: (c)
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