JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2025
- Medium
- Numerical
A loop ABCDA , carrying current \(\mathrm{I}=12 \mathrm{~A}\), is placed in a plane, consists of two semi-circular segments of radius \(R_1=6 \pi \mathrm{~m}\) and \(\mathrm{R}_2=4 \pi \mathrm{~m}\). The magnitude of the resultant magnetic field at center O is \(\mathrm{k} \times 10^{-7} \mathrm{~T}\). The value of k is ______ (Given \(\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}\))

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Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
Magnetic field due to \(\mathrm{AB} \& \mathrm{CD}=0\)
\(\begin{aligned}
& \mathrm{B}_0=\left|\mathrm{B}_{\mathrm{R}_1}-\mathrm{B}_{\mathrm{R}_2}\right| \\
& =\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}_2}-\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}_1} \\
& =\frac{4 \pi \times 10^{-7} \times 12}{4}\left(\frac{1}{4 \pi}-\frac{1}{6 \pi}\right) \\
& =12 \pi \times 10^{-7}\left(\frac{1}{12 \pi}\right) \\
& =1 \times 10^{-7}
\end{aligned}\)
\(K=1\)
\(\begin{aligned}
& \mathrm{B}_0=\left|\mathrm{B}_{\mathrm{R}_1}-\mathrm{B}_{\mathrm{R}_2}\right| \\
& =\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}_2}-\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}_1} \\
& =\frac{4 \pi \times 10^{-7} \times 12}{4}\left(\frac{1}{4 \pi}-\frac{1}{6 \pi}\right) \\
& =12 \pi \times 10^{-7}\left(\frac{1}{12 \pi}\right) \\
& =1 \times 10^{-7}
\end{aligned}\)
\(K=1\)
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