MHT CET · Physics · Magnetic Effects of Current
The magnetic field at the centre of a current carrying circular coil of area ' \(\mathrm{A}\) ' is ' \(\mathrm{B}\) '. The magnetic moment of the coil is ( \(\mu_0=\) permeability of free space)
- A \(\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \sqrt{\pi}}\)
- B \(\frac{\mathrm{BA}^{3 / 2}}{\mu_0 \pi}\)
- C \(\frac{\mu_0 \sqrt{\pi}}{2 \mathrm{BA}^2}\)
- D \(\frac{2 \mathrm{BA}^2}{\mu_0 \sqrt{\pi}}\)
Answer & Solution
Correct Answer
(A) \(\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \sqrt{\pi}}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& A=\pi R^2 \\
& \Rightarrow R=\sqrt{\frac{A}{\pi}} \\
& B=\frac{\mu_0 I}{2 R}
\end{aligned}
\)
Magnetic Moment,
\(
\begin{aligned}
& M=I A=\frac{2 A B R}{\mu_0} \\
& =\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \sqrt{\pi}}
\end{aligned}
\)
\begin{aligned}
& A=\pi R^2 \\
& \Rightarrow R=\sqrt{\frac{A}{\pi}} \\
& B=\frac{\mu_0 I}{2 R}
\end{aligned}
\)
Magnetic Moment,
\(
\begin{aligned}
& M=I A=\frac{2 A B R}{\mu_0} \\
& =\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \sqrt{\pi}}
\end{aligned}
\)
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