Question

A square coil of side $l$ carries a current $I$. It is kept on a smooth horizontal surface. A uniform magnetic field exists in the region, directed in the plane of the coil as shown in the figure. what should be the minimum value of $B$, so that the coil will start tipping over ? Mass of the coil is $M$ and it has $N$ number of turns.

• A. $\frac{mg}{2NIl}$
• B. $\frac{Mg}{NIl}$
• C. $\frac{Mg}{NI{l}^2}$
• D. $\frac{Mg}{2NI{l}^2}$

Answer 1

The correct option is A $\frac{mg}{2NIl}$

The forces acting on sides $\text{QR}$ and $\text{SP}$ are each equal to zero while the force acting on $\text{PQ}$ and $\text{RS}$ are as shown below,


Now, the weight of the loop will have torque in clockwise direction and torque due to magnetic field will be in anti-clock wise direction, at the instant when the coil is just about to tip over.

${\tau }_m={\tau }_{mg}$

$\therefore NBIA=mg.\frac{l}{2}$

$NBI.l^2=mg\frac{l}{2}$

$\therefore B=\frac{mg}{2NIl}$

Hence, option $\left(A\right)$ is the correct answer.

Why this Question? Tip: Torque due to a current carrying coil placed in a uniform magnetic field is given by, $\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$ $\tau =BINA\sin \theta$