MHT CET · Physics · Magnetic Effects of Current
The magnetic field at the centre of a current carrying circular coil of area ' \(A\) ' is ' \(B\) '. The magnetic moment of the coil is ( \(\mu_0=\) permeability of free space)
- A \(\frac{2 \mu_0 \pi^{1 / 2}}{\mathrm{BA}^{3 / 2}}\)
- B \(\frac{\mathrm{BA}^{3 / 2}}{\mu_0 \pi}\)
- C \(\frac{2 B \mathrm{~A}^{3 / 2}}{\mu_0 \pi^{1 / 2}}\)
- D \(\frac{\mathrm{BA}^2}{\mu_0 \pi}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 B \mathrm{~A}^{3 / 2}}{\mu_0 \pi^{1 / 2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& A=\pi r^2 \Rightarrow r=\sqrt{\frac{A}{\pi}} \\
\therefore \quad & B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 I}{2 \sqrt{\frac{A}{\pi}}} \\
\therefore \quad & I=\frac{2 B}{\mu_0} \sqrt{\frac{A}{\pi}}
\end{array}\)
Magnetic moment \(\mathrm{m}=\mathrm{IA}\)
\(\mathrm{IA}=\frac{2 \mathrm{~B}}{\mu_0} \sqrt{\frac{\mathrm{~A}}{\pi}} \times \mathrm{A}=\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \pi^{1 / 2}}\)
& A=\pi r^2 \Rightarrow r=\sqrt{\frac{A}{\pi}} \\
\therefore \quad & B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 I}{2 \sqrt{\frac{A}{\pi}}} \\
\therefore \quad & I=\frac{2 B}{\mu_0} \sqrt{\frac{A}{\pi}}
\end{array}\)
Magnetic moment \(\mathrm{m}=\mathrm{IA}\)
\(\mathrm{IA}=\frac{2 \mathrm{~B}}{\mu_0} \sqrt{\frac{\mathrm{~A}}{\pi}} \times \mathrm{A}=\frac{2 \mathrm{BA}^{3 / 2}}{\mu_0 \pi^{1 / 2}}\)
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