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 electromagnet as shown in fig. The area of the coil is $S=1c{m}^2$, the length of the right arm 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, equilibrium is restored by putting an additional weight of mass $m=60$ mg on the balance pan. Find the magnetic induction field (in terms of $\times {10}^{-1}T)$ between the poles of the electromagnet, assuming it to be uniform :

Answer 1

Torque on the loop due to magnetic field ${\tau }_m=NISB\sin {90}^{\circ }$
Torque due to extra weight added ${\tau }_g=\left(\Delta M\right)gl$
Equating both torques, $NISB\sin {90}^{\circ }=\left(\Delta M\right)gl$
$\therefore B=\frac{\left(\Delta M\right)gl}{NIS}=\frac{(60\times {10}^{-6})\times 9.8\times (30\times {10}^{-2})}{200\times (22\times {10}^{-3})}=0.4T=4\times {10}^{-1}T$
Hence, the magnetic field in units of ${10}^{-1}T$ is 4.