Q113 / 150 · MHT CET 2020 (12 Oct Shift 1) · Physics

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
    $\frac{B A^{\frac{3}{2}}}{\mu_{0} \pi}$
  • Correct answer
    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
    $\frac{\mathrm{B} A^{\frac{3}{2}}}{\mu_{0} \pi^{\frac{1}{2}}}$

Solution

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 $$ \begin{array}{c} \mathrm{M}=\mathrm{IA}=\frac{2 \mathrm{~B}}{\mu_{0}} \cdot \sqrt{\frac{\mathrm{A}}{\pi}} \cdot \mathrm{A} \\ =\frac{2 \mathrm{BA}^{3 / 2}}{\mu \pi^{\frac{1}{2}}} \end{array} $$

Hence, the correct answer is option: 'B'.

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