Question

A circular loop of mass m and radius r is lying in a horizontal (xy-plane) table as shown in figure. A uniform magnetic filed B is applied parallel to X-axis. The current I in the loop, so that its one edge just lifts from the table, is

• A. $\frac{mg}{{\pi r}^{2}B}$
• B. $\frac{mg}{\pi rB}$
• C. $\frac{mg}{2\pi rB}$
• D. $\frac{\pi rB}{mgl}$

Answer 1

The correct option is C $\frac{mg}{2\pi rB}$

Given :

$\overrightarrow{B}=B\hat{x}$

Current in the loop = $I$

Mass of wire = $m$

Radius of wire = $r$

Solution :

We know that Force due to the magnetic field is given as

$F=I\int (dl\times B)$

Note that $\int dl=2\pi r\hat{z}$ (Circumference of circle)

Now, $\int (dl\times B)=2\pi rB\hat{y}$

Finally $F=2\pi rBI\hat{y}$

Note that the gravitational force acts downwards and the magnetic force acts upwards.

For lifting its one edge the magnitude of magnetic force should balance the magnitude of the gravitational force.

Hence, $mg=2\pi rBI$

Hence, $I=\frac{mg}{2\pi rB}$

The Correct Opt : [C]