Question

A rectangular coil of conductor ABCD with the length $AD=3$ and breadth $DC=2m$is hung from one of the arms of a balance, as shown in the Figure. A current $I=8A$ is passed through the coil. When the magnetic Field turned off, a mass M kg is added to the pan of other arm to balance the coil. When the magnetic field of strength $0.4T$ is switched on then, the system is

• A. in equilibrium by taking a mass of 0.32 kg from left pan.
• B. in equilibrium by adding a mass of 0.64 kg to the left pan.
• C. in equilibrium by reversing the current and taking a mass of 0.64kg from the left pan.
• D. in equilibrium by reversing the current and adding a mass of 0.32kg to the left pan.

Answer 1

Answer: (B), (C)

When the magnetic field is switched off, for equilibrium, if W is the weight of the coil, then, net torque acting about O should be zero.
When the field is switched on, the force on CD$=ILB$ downward. Hence to balance, mass m should be added to the left pan, such that,
$mg=ILB$ or $m=\frac{ILB}{g}$
$\therefore m=\frac{8\times 2\times 0.4}{10}=0.64kg$
By reversing the current, a mass of 0.64kg should be taken from the left pan.