A uniform circular loop of radius a and resistance $\mathrm{R}$ is pulled at a constant velocity $\mathrm{v}$ out of a region of a uniform magnetic field whose magnitude is $\mathrm{B}$. The plane of the loop and the velocity are both perpendicular to $\mathrm{B}$. Then the electrical power in the circular loop at the instant when the arc (of the circular loop) outside the region of magnetic field subtends an angle $\frac{\mathrm{\pi }}{3}$ at the centre of the loop is

• Correct answer
  A

  $\frac{{\mathrm{B}}^2{\mathrm{a}}^2{\mathrm{v}}^2}{\mathrm{R}}$
• B

  $\frac{2{\mathrm{B}}^2{\mathrm{a}}^2{\mathrm{v}}^2}{\mathrm{R}}$
• C

  $\frac{{\mathrm{B}}^2{\mathrm{a}}^2{\mathrm{v}}^2}{2\mathrm{R}}$
• D

  none of these

Solution

The induced emf in the circular loop $=\mathrm{B}\left(\mathrm{PQ}\right)\mathrm{v}=\mathrm{Bav}$

Therefore electrical power consumed by the circular loop

$=\frac{{\left(\mathrm{emf}\right)}^2}{\mathrm{R}}=\frac{{\left(\mathrm{Bav}\right)}^2}{\mathrm{R}}$

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