A current of 10 A is flowing in a wire of length 1.5 m. When placed in a uniform magnetic field of 2 T, it experiences a force 15 N. The angle between the magnetic field and the direction of flow of current is

30°

The correct answer is Option (2) → 30°

Given:

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

Length of wire: $L = 1.5~\text{m}$

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

Force on wire: $F = 15~\text{N}$

Force on a current-carrying wire in a magnetic field:

$F = I L B \sin \theta$

Substitute values:

$15 = 10 \cdot 1.5 \cdot 2 \cdot \sin \theta$

$15 = 30 \sin \theta \Rightarrow \sin \theta = \frac{15}{30} = 0.5$

Answer: $\theta = 30^\circ$