Q.
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 figure. The cross sectional area of coil is $A = 1.0\ cm^{2}$, length of arm OA of the balance beam is $l = 30\ cm$. When there is no current in the coil the balance is in equilibrium. On passing a current $I = 22$ mA through the coil the equilibrium is restored by putting the additional counter weight of mass $\mathrm{\Delta}m = 60\ mg$ on the balance pan. Find the magnetic induction at the spot where coil is located
A) 0.4 $T$
B) 0.3 $T$
C) 0.2 $T$
D) 0.1 $T$
Explanation
(a) On passing current through the coil, it acts as a magnetic dipole. Torque acting on magnetic dipole is counter balanced by the moment of additional weight about position $O$. Torque acting on a magnetic dipole
$\tau = MB\sin\theta = (NiA)B\sin{90{^\circ}} = NiAB$
Again $\tau = Force \times Lever\ arm = \mathrm{\Delta}mg \times l$
$\Rightarrow NiAB = \mathrm{\Delta}mgl;\ \mathrm{\Delta}m = 60 \times 10^{- 3}g = 60 \times 10^{- 6}\ kg$
$\Rightarrow B = \frac{\mathrm{\Delta}mgl}{NiA} = \frac{60 \times 10^{- 6} \times 9.8 \times 30 \times 10^{- 2}}{200 \times 22 \times 10^{- 3} \times 1 \times 10^{- 4}} = 0.4\ T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.