MHT CET · Physics · Magnetic Effects of Current
Magnetic field at the centre of a circular loop of area ' \(A\) ' is ' \(B\) '. The magnetic moment of the loop will be
- A \(\frac{\mathrm{BA}^2}{\mu_0 \pi}\)
- B \(\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \pi}\)
- C \(\frac{\mathrm{BA}^{1 / 2}}{\mu_0 \pi^{1 / 2}}\)
- D \(\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \pi^{1 / 2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \pi^{1 / 2}}\)
Step-by-step Solution
Detailed explanation
Magnetic field at centre of circular loop is given by,
\(B=\frac{\mu_0 I}{2 r} \Rightarrow I=\frac{2 B r}{\mu_0}...(i)\)
For circular loop,
\(\mathrm{A}=\pi \mathrm{r}^2 \mathrm{r}=\sqrt{\frac{\mathrm{A}}{\pi}}...(ii)\)
Magnetic moment of the loop will be,
\(\begin{aligned}
& \mathrm{M}=\mathrm{IA} \\
& \mathrm{M}=\frac{2 \mathrm{Br}}{\mu_0} \times \mathrm{A}...[From(i)] \\
& \mathrm{M}=\frac{2 \mathrm{BA}}{\mu_0} \times \sqrt{\frac{\mathrm{A}}{\pi}}...[From(ii)] \\
& \mathrm{M}=\frac{2 \mathrm{BA}^{\frac{3}{2}}}{\mu_0(\pi)^{\frac{1}{2}}}
\end{aligned}\)
\(B=\frac{\mu_0 I}{2 r} \Rightarrow I=\frac{2 B r}{\mu_0}...(i)\)
For circular loop,
\(\mathrm{A}=\pi \mathrm{r}^2 \mathrm{r}=\sqrt{\frac{\mathrm{A}}{\pi}}...(ii)\)
Magnetic moment of the loop will be,
\(\begin{aligned}
& \mathrm{M}=\mathrm{IA} \\
& \mathrm{M}=\frac{2 \mathrm{Br}}{\mu_0} \times \mathrm{A}...[From(i)] \\
& \mathrm{M}=\frac{2 \mathrm{BA}}{\mu_0} \times \sqrt{\frac{\mathrm{A}}{\pi}}...[From(ii)] \\
& \mathrm{M}=\frac{2 \mathrm{BA}^{\frac{3}{2}}}{\mu_0(\pi)^{\frac{1}{2}}}
\end{aligned}\)
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