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

  $2 \pi i r \mathrm{~B}_{0}$
• 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 $\mathrm{AB}$ the direction of force will be Leftwards and the magnitude will be $$ \mathrm{dF}=\operatorname{Idl} \mathrm{B} \sin 90^{\circ}=\mathrm{IdIB} $$

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