MHT CET · Physics · Magnetic Effects of Current
Magnetic field at the centre of a circular loop of area 'A' is ' \(B^{\prime}\) '. The magnetic moment of the loop will be \(\left(\mu_{0}=\right.\) permeability of free space)
- A \(\frac{B A^{\frac{3}{2}}}{\mu_{0} \pi}\)
- B \(\frac{2 \mathrm{~B} A^{\frac{3}{2}}}{\mu_{0} \pi^{\frac{1}{2}}}\)
- C \(\frac{2 \mathrm{BA}^{2}}{\mu_{0} \pi}\)
- D \(\frac{\mathrm{B} A^{\frac{3}{2}}}{\mu_{0} \pi^{\frac{1}{2}}}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 \mathrm{~B} A^{\frac{3}{2}}}{\mu_{0} \pi^{\frac{1}{2}}}\)
Step-by-step Solution
Detailed explanation
If \(\mathrm{r}\) is the radius of the circular loop them the area of the loop \(\mathrm{A}=\pi \mathrm{r}^{2}\) \(\therefore \mathrm{r}=\sqrt{\frac{\mathrm{A}}{\pi}}\)
\(\therefore\) The magnetic field at the centre is given by
\(
\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}=\frac{\mu_{0} \mathrm{I}}{2 \sqrt{\frac{\mathrm{A}}{\pi}}}
\)
\(
\therefore I=\frac{2 B}{\mu_{0}} \cdot \sqrt{\frac{A}{\pi}}
\)
Magnetic moment
\(
\begin{array}{c}
\mathrm{M}=\mathrm{IA}=\frac{2 \mathrm{~B}}{\mu_{0}} \cdot \sqrt{\frac{\mathrm{A}}{\pi}} \cdot \mathrm{A} \\
=\frac{2 \mathrm{BA}^{3 / 2}}{\mu \pi^{\frac{1}{2}}}
\end{array}
\)
\(\therefore\) The magnetic field at the centre is given by
\(
\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}=\frac{\mu_{0} \mathrm{I}}{2 \sqrt{\frac{\mathrm{A}}{\pi}}}
\)
\(
\therefore I=\frac{2 B}{\mu_{0}} \cdot \sqrt{\frac{A}{\pi}}
\)
Magnetic moment
\(
\begin{array}{c}
\mathrm{M}=\mathrm{IA}=\frac{2 \mathrm{~B}}{\mu_{0}} \cdot \sqrt{\frac{\mathrm{A}}{\pi}} \cdot \mathrm{A} \\
=\frac{2 \mathrm{BA}^{3 / 2}}{\mu \pi^{\frac{1}{2}}}
\end{array}
\)
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