A loop A B C D A , carrying current I=12 A , is placed in a plane, consists of two semi-circular segments of radius R 1 =6 π m and R 2 =4 π m . The magnitude of the resultant magnetic field at center O is k × 10 -7 T . The value of k is . (Given μ 0 =4 π × 10 -7 Tm A -1 )

Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 3 April, Morning Shift — Question 63

A loop $A B C D A$ , carrying current $I=12 \mathrm{~A}$ , is placed in a plane,

consists of two semi-circular segments of radius

$R_{1}=6 \pi \mathrm{~m}$ and $R_{2}=4 \pi \mathrm{~m}$ .

The magnitude of the resultant magnetic field at center $O$ is $k \times 10^{-7} \mathrm{~T}$ .

The value of $k$ is $\qquad$ .

(Given $\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$ )

Numerical answer — enter this value.

$B=\frac{\mu_{0}}{2 r} \frac{\theta}{2 \pi}$ for an arc For semicircle $B=\frac{\mu_{0} l}{4 r}$

\begin{aligned} B_{\text {net }} & =\frac{\mu_{0} I}{4\{4 \pi\}}-\frac{\mu_{0} I}{4\{6 \pi\}} \\ & =\frac{4_{0} I}{4 \pi}\left\{\frac{1}{4}-\frac{1}{6}\right\} \\ & =10^{-7} \times 12 \times \frac{2}{24}=10^{-7} \mathrm{~T} \Rightarrow k=1 \end{aligned}

Answer key and solution verified before publishing.

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