As shown in the figure, a long straight conductor with semicircular arc of radius π/10 m is carrying current I = 3 A. The magnitude of the magnetic field, at the center O of the arc is: (The permeability of the vacuum = 4π × 10⁻⁷ NA⁻²) (a) 4 (b) 3 (c) 6 (d) 1

Answer: (b). Both straight parts lie on a line through O, so they give zero field. Semicircle: B = (μ₀ I)/(4R) B = 4π × 10⁻⁷ × 34 × π/10 = 3 × 10⁻⁶ T So the answer is (b) 3 μ T.

As shown in the figure, a long straight conductor with semicircular… | JEE Main 2023 (1 Feb, Shift 2)

Q42JEE Main 2023 · 1 FEB · SHIFT 2Medium

Magnetic Field of Straight Wires

As shown in the figure, a long straight conductor with semicircular arc of radius $\frac{\pi}{10}$ m is carrying current $I = 3\text{ A}$ . The magnitude of the magnetic field, at the center O of the arc is: (The permeability of the vacuum $= 4\pi \times 10^{-7}\text{ NA}^{-2}$ )

Show answer and solution

Answer: (b)

Both straight parts lie on a line through O, so they give zero field.
Semicircle: $B = \frac{\mu_0 I}{4R}$
$B = \frac{4\pi \times 10^{-7} \times 3}{4 \times \pi/10} = 3 \times 10^{-6}$ T
So the answer is (b) $3\ \mu$ T.

Report a mistake

Magnetic Field of Straight Wires question from JEE Main 2023 (1 Feb, Shift 2)

More questions on Magnetic Field of Straight Wires