A current of 10 A is flowing in a wire of length 1.2 m. A force of 12 N acts on it when it is placed in a uniform magnetic field of 2 T. The angle between the magnetic field and the direction of the current is

30°

The correct answer is Option (1) → 30°

Given:

Current: $I = 10\ \text{A}$

Length of wire: $L = 1.2\ \text{m}$

Force: $F = 12\ \text{N}$

Magnetic field: $B = 2\ \text{T}$

Magnetic force on a current-carrying wire: $F = I L B \sin\theta$

Solving for $\theta$:

$\sin\theta = \frac{F}{I L B} = \frac{12}{10 \cdot 1.2 \cdot 2} = \frac{12}{24} = 0.5$

$\theta = \arcsin(0.5) = 30^\circ$

∴ Angle between the magnetic field and current = 30°