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
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 30^\circ \)
The magnetic force on a current-carrying wire is \( F = I l B\sin\theta \), where \( \theta \) is the angle between the wire (current direction) and the field.
\( \displaystyle \sin\theta = \frac{F}{I l B} = \frac{15}{10 \times 1.5 \times 2} = \frac{15}{30} = 0.5 \)
\( \displaystyle \theta = 30^\circ \)
Note: A similar question was also asked in CUET (UG) 2026 on 15-05-2026 (Slot 1).