A circular loop of area \( 0.01\,\mathrm{m^2} \) carrying a current of 10 A, is held in a magnetic field of 0.1 T parallel to the plane of the loop. The torque acting on the loop is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.01\,\mathrm{N\,m} \)
The torque on a current loop is \( \tau = mB\sin\theta \), where \( m = NIA \) is its magnetic moment, directed along the normal to the loop. A field parallel to the plane of the loop is perpendicular to the normal, so \( \theta = 90^\circ \) and the torque is maximum.
\( N = 1 \)
\( I = 10\,\mathrm{A} \)
\( A = 0.01\,\mathrm{m^2} \)
\( B = 0.1\,\mathrm{T} \)
\( \displaystyle \tau = NIAB\sin90^\circ = 1 \times 10 \times 0.01 \times 0.1 \times 1 = 0.01\,\mathrm{N\,m} \)