A rectangular coil of 20 turns has sides 10 cm and 5 cm. It carries a current of 2 A. It is placed in a uniform magnetic field of 0.5 T. The field is in the plane of the coil and parallel to its longer sides. The magnitude of the force on each shorter side of the coil is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(1\ \mathrm{N}\)
A side that is parallel to the field gets no force. Here the longer sides are parallel to the field, so only the shorter sides get a force.
Each shorter side is perpendicular to the field. In one turn it gets a force \( IbB \), where \( b \) is its length. With \( N \) turns, the force on the side is \( F=NIbB \).
The forces on the two shorter sides are equal and opposite. They give no net force, but they turn the coil.
Here \( N=20 \), \( I=2\ \mathrm{A} \), \( b=5\ \mathrm{cm}=0.05\ \mathrm{m} \) and \( B=0.5\ \mathrm{T} \).
\( \begin{aligned} F &= NIbB \\ &= 20\times2\times0.05\times0.5 \\ &= 1\ \mathrm{N} \end{aligned} \)
NCERT (2026-27):
Torque on a rectangular current loop in a uniform magnetic field – “The field exerts no force on the two arms AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force F1 on it which is directed into the plane of the loop. Its magnitude is,”
\( F_1=IbB \)