A conducting circular loop of radius $r$ carries a constant current $i .$ It is placed in a uniform magnetic field $\overrightarrow{\mathrm{B}}_{0}$ such that $\overrightarrow{\mathrm{B}}_{0}$ is perpendicular to the plane of the loop. The magnetic force acting on the loop is

• A

  $i r \mathrm{~B}_{0}$
• B

  $2 \pi \operatorname{ir} \mathrm{B}_{0}$
• Correct answer
  C

  zero
• D

  $\pi$ ir $\mathrm{B}_{0}$

Solution

The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Fleming's Left hand rule on element AB the direction of force will be Leftwards and the magnitude will be $d F=\mid d l B \sin 90^{\circ}=I d l B$
On element $\mathrm{CD}$, the direction of force will be towards right on the plane of the papper and the magnitude will be $\mathrm{dF}=\mathrm{Id} / \mathrm{B}$.

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