MHT CET · Physics · Magnetic Effects of Current
Magnetic induction produced at the centre of a circular loop of radius ' R ' carrying a current is ' \(B\) '. The magnetic moment of the loop is ( \(\mu_0=\) permeability of free space)
- A \(\frac{\mathrm{BR}^3}{2 \pi \mu_0}\)
- B \(\frac{2 \pi \mathrm{BR}^3}{\mu_0}\)
- C \(\frac{\mathrm{BR}^2}{2 \pi \mu_0}\)
- D \(\frac{2 \pi \mathrm{BR}^2}{\dot{\mu}_0}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 \pi \mathrm{BR}^3}{\mu_0}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{m}=\mathrm{nIA}\)
For coil, magnetic induction at the centre,
\(\begin{aligned}
& \quad B=\frac{\mu_0 n I}{2 R} \\
& \therefore \quad I=\frac{B \times 2 R}{\mu_0 n} \\
& \\
& \text { For } n=1, \text { Area } A=\pi R^2 \\
& \\
& m=\frac{B \times 2 R}{\mu_0} \times \pi R^2=\frac{2 \pi B R^3}{\mu_0}
\end{aligned}\)
For coil, magnetic induction at the centre,
\(\begin{aligned}
& \quad B=\frac{\mu_0 n I}{2 R} \\
& \therefore \quad I=\frac{B \times 2 R}{\mu_0 n} \\
& \\
& \text { For } n=1, \text { Area } A=\pi R^2 \\
& \\
& m=\frac{B \times 2 R}{\mu_0} \times \pi R^2=\frac{2 \pi B R^3}{\mu_0}
\end{aligned}\)
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