As shown in the figure, a long straight conductor with semicircular arc of radius \frac\pi10 m is carrying current I=3A . The magnitude of the magnetic field, at the center O of the arc is : (The permeability of the vacuum =4\pi\times10^-7~NA^-2 )

Magnetics PYQ — JEE Main 2023 · 1 Feb · Shift 2 · Q60 with Solution

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Correct answer: B

Magnetic field at the center due to the semicircular arc

For a circular arc of angle $\theta$ (in radians), the magnetic field at the center is

B_{\text{arc}}=\frac{\mu_0 I\theta}{4\pi R}

For a semicircle, $\theta=\pi$ , so

B_{\text{semi}}=\frac{\mu_0 I}{4R}

Given:

\mu_0=4\pi\times10^{-7}\,\text{N A}^{-2},\quad I=3\,\text{A},\quad R=\frac{\pi}{10}\,\text{m}

Thus,

B_{\text{semi}}=\frac{4\pi\times10^{-7}\times 3}{4\times (\pi/10)}

=\frac{12\pi\times10^{-7}}{4\pi/10} =12\pi\times10^{-7}\times \frac{10}{4\pi} =30\times10^{-7}

B_{\text{semi}}=3\times10^{-6}\,\text{T}=3\,\mu\text{T}

Magnetic field due to the straight portions

From the usual geometry of this standard figure, the straight portions lie along the same line passing through the center $O$ . For each current element on these straight parts, the position vector toward $O$ is parallel to $d\vec l$ , hence

d\vec B\propto d\vec l\times \hat r =0

So, the straight conductor contributes zero magnetic field at $O$ .

Net magnetic field at $O$

Therefore,

B_{\text{net}}=B_{\text{semi}}=3\,\mu\text{T}

Option check

A: $4\mu\text{T}$ ❌
B: $3\mu\text{T}$ ✅
C: $6\mu\text{T}$ ❌
D: $1\mu\text{T}$ ❌

Hence, the correct answer is Option B.

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