A conducting circular loop of radius $ \mathrm{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

Question

A conducting circular loop of radius $ \mathrm{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, ir $ \overrightarrow{\mathrm{B}_{0}} $, $ 2 \pi $ ir $ \overrightarrow{\mathrm{B}_{0}} $, Zero, $ \pi $ ir $ \overrightarrow{\mathrm{B}_{0}} $

MARKS App · Accepted Answer

The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements on the loop. Applying Fleming’s left-hand rule on element A B , the direction of force will be leftwards and the magnitude will be d F = I d l B sin 90 o = I d l B On element C D , the direction of force will be rightwards on the plane of the paper and the magnitude will be d F = I d l B . Hence, the net force on the loop due to current elements will be zero, which implies that the total force on the loop will be zero.

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