MHT CET · Physics · Magnetic Effects of Current
Figure shows two semicircular loops of radii \(R_1\) and \(R_2\) carrying current \(I\). The magnetic field at the common centre ' \(\mathrm{O}\) ' is
- A \(\frac{\mu_0 I}{4}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\)
- B \(\frac{\mu_0 I}{4}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
- C \(\frac{\mu_0 I}{2 \pi}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\)
- D \(\frac{\mu_0 I}{2 \pi}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
Answer & Solution
Correct Answer
(A) \(\frac{\mu_0 I}{4}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\)
Step-by-step Solution
Detailed explanation
For semicircular arc the magnetic field is given as \(B=\frac{\mu_0 i}{4 R}\)
The equivalent magnetic field at the centre is
\(\begin{aligned}
& B_{\text {eq }}=\frac{\mu_0 I}{4 R_1}+\frac{\mu_0 I}{4 R_2} \\
& B_{\text {eq }}=\frac{\mu_0 I}{4}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)
\end{aligned}\)
The equivalent magnetic field at the centre is
\(\begin{aligned}
& B_{\text {eq }}=\frac{\mu_0 I}{4 R_1}+\frac{\mu_0 I}{4 R_2} \\
& B_{\text {eq }}=\frac{\mu_0 I}{4}\left(\frac{1}{R_1}+\frac{1}{R_2}\right)
\end{aligned}\)
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