Question

A small coil $C$ with $N=200$ turns is mounted on one end of a balance beam and introduced between the poles of an electro-magnet as shown in figure above. The cross-sectional area of the coil is $S=1.0c{m}^2$, the length of the arm $OA$ of the balance beam is $l=30cm$. When there is no current in the coil the balance is in equilibrium. On passing a current $I=22mA$ through the coil the equilibrium is restored by putting the additional counterweight of mass $\Delta m=60mg$ on the balance pan. The magnetic induction at the spot where the coil is located is given as $\frac{x}{10}T$

Answer 1

We know that the torque acting on a magnetic dipole,
$\overrightarrow{N}=\overrightarrow{p_m}\times \overrightarrow{B}$
But, $\overrightarrow{p_m}=iS\hat{n}$, where $\hat{n}$ is the normal on the plane of the loop and is directed in the direction of advancement of a right handed screw, if we rotate the screw in the sense of current in the loop.
On passing a current through the coil, this torque acting on the magnetic dipole, is counter-balanced by the moment of additional weight, about $O$. Hence, the direction of current in the loop must be in the direction, shown in figure below.
$\overrightarrow{p_m}\times \overrightarrow{B}=-\overrightarrow{l}\times \Delta m\overrightarrow{g}$
or, $NiSB=\Delta mgl$
So, $B=\frac{\Delta mgl}{NiS}=0.4T$ on putting the values.