Question

A coil of area $5 \, \text{cm}^2$ is placed in a uniform magnetic field of $1.5 \, \text{m}$. If the coil has 100 turns and $0.2 \, \text{A}$ of current is passed in it, then find:

Magnetic dipole moment of the coil
Maximum torque on the coil

Hint:

Magnetic dipole moment depends on $NIA$, while torque depends on alignment with the magnetic field.

Answer 1

Step 1: Formula for magnetic dipole moment.
\[ m = N \cdot I \cdot A \] where $N =$ number of turns, $I =$ current, $A =$ area.
Step 2: Substitution.
Given: \[ A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2, \quad N = 100, \quad I = 0.2 \, A. \] \[ m = 100 \times 0.2 \times 5 \times 10^{-4}. \] \[ m = 1.0 \times 10^{-3} \, \text{Am}^2. \]
Step 3: Formula for torque.
\[ \tau = m B \sin \theta \] For maximum torque, $\sin \theta = 1$: \[ \tau_{max} = m B. \]
Step 4: Substitution.
\[ \tau_{max} = (1.0 \times 10^{-3})(1.5) = 1.5 \times 10^{-3} \, \text{Nm}. \]
Step 5: Conclusion.
1. $m = 1 \times 10^{-3} \, \text{Am}^2$
2. $\tau_{max} = 1.5 \times 10^{-3} \, \text{Nm}$