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 figure. The cross sectional area of coil is $A=1.0c{m}^2$, length of arm OA of the balance beam is 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 $\Delta m=60mg$ 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

Answer 1

The correct option is A 0.4 T

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
$\tau =MBsin\theta =\left(NiA\right)Bsin{90}^{\circ }=NiABAgain\tau =Force\times Leverarm=\Delta mg\times l\Rightarrow NiAB=\Delta mgl\Rightarrow B=\frac{\Delta mgl}{NiA}=\frac{60\times {10}^{-3}\times 9.8\times 30\times {10}^{-2}}{200\times 22\times {10}^{-3}\times 1\times {10}^{-4}}=0.4T$