Question

A coil having 100 turns, area of $5 \times 10^{-3} \, \text{m}^2$, carrying current of $1 \, \text{mA}$ is placed in a uniform magnetic field of $0.20 \, \text{T}$ such a way that the plane of the coil is perpendicular to the magnetic field.
The work done in turning the coil through $90^\circ$ is ____ $\mu \text{J}$.

Answer 1

Correct Answer: 100

The work done \( W = \Delta U = U_f - U_i \):
\[W = -(\mu B)_f - (-\mu B)_i\]
Initially, the magnetic moment \( \mu \) is perpendicular to the magnetic field, so:
\[W = 0 + (\mu B)\]
Substitute the values:
\[\mu = (100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}) \, \text{A} \cdot \text{m}^2\]
\[W = (1 \times 10^{-4}) \times 0.2 \, \text{J} = 1 \times 10^{-5} \, \text{J} = 100 \, \mu \text{J}\]