A loop ABCDA , carrying current I=12 ~A , is placed in a plane, consists of two semi-circular segments of radius R_1=6 \pi ~m and R_2=4 \pi ~m . The magnitude of the resultant magnetic field at center O is k \times 10^-7 ~T . The value of k is \underline\hspace2cm . ( Given \mu_0=4 \pi \times 10^-7 Tm A^-1 )

Magnetics PYQ — JEE Main 2025 · 3 Apr · Shift 1 · Q72 with Solution

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Correct answer: 1

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

B=\frac{\mu_0 I\theta}{4\pi R}

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

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

The two straight connecting segments lie along lines passing through the center $O$ . For each current element on such a segment,

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

Since $d\vec l$ is parallel to $\hat r$ , their cross product is zero. Hence, these straight segments produce no magnetic field at $O$ .

The loop consists of two semicircular arcs of radii:

R_1=6\pi\,\text{m}, \qquad R_2=4\pi\,\text{m}

Their currents around the center are in opposite senses, so the magnetic fields at $O$ oppose each other.

Thus,

B_{\text{net}}=\left|\frac{\mu_0 I}{4R_2}-\frac{\mu_0 I}{4R_1}\right|

Substitute values:

B_{\text{net}}=\frac{\mu_0 I}{4}\left(\frac{1}{R_2}-\frac{1}{R_1}\right)

Using $\mu_0=4\pi\times10^{-7}$ and $I=12$ A,

B_{\text{net}}=\frac{4\pi\times10^{-7}\times 12}{4}\left(\frac{1}{4\pi}-\frac{1}{6\pi}\right)

=12\pi\times10^{-7}\left(\frac{3-2}{12\pi}\right)

=12\pi\times10^{-7}\cdot \frac{1}{12\pi}=10^{-7}\,\text{T}

Given,

B_{\text{net}}=k\times10^{-7}\,\text{T}

So,

k=1

Stored correct answer: $1$

My derived answer matches the stored answer.