A particle of charge 16 × 10⁻¹⁶ C moving with velocity 10 ms ⁻¹ along x-axis enters a region where magnetic field of induction B is along the y-axis and an electric field of magnitude 10⁴ Vm ⁻¹ is along the negative z-axis. If the charged particle continues moving along x-axis, the magnitude of B is: (a) 16 × 10³ Wb m ⁻² (b) 2 × 10³ Wb m ⁻² (c) 1 × 10³ Wb m ⁻² (d) 4 × 10³ Wb m ⁻²

Answer: (c). Undeflected motion ⇒ qE = qvB B = E/v = 10⁴10 = 10³ Wb m⁻² So the answer is (c) 1×10³ Wb m⁻².

A particle of charge 16 × 10⁻¹⁶ C moving with velocity 10 ms ⁻¹ along… | JEE Main 2013 (23 Apr, Shift 1)

Q101JEE Main 2013 · 23 APR · SHIFT 1Easy

Moving Charge in a Magnetic Field

A particle of charge $16 \times 10^{-16}$ C moving with velocity 10 ms ${}^{-1}$ along x-axis enters a region where magnetic field of induction $\vec{B}$ is along the y-axis and an electric field of magnitude $10^{4}$ Vm ${}^{-1}$ is along the negative z-axis. If the charged particle continues moving along x-axis, the magnitude of $\vec{B}$ is:

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Answer: (c)

Undeflected motion $\Rightarrow qE = qvB$
$B = \dfrac{E}{v} = \dfrac{10^{4}}{10} = 10^{3}\text{ Wb m}^{-2}$
So the answer is (c) $1\times10^{3}\text{ Wb m}^{-2}$ .

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