A square loop with each side 1 cm, carrying a current of 10 A, is placed in a magnetic field of 0.2 T. The direction of magnetic field is parallel to the plane of the loop. The torque experienced by the loop is _____.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\times10^{-4}\,\mathrm{Nm} \)
The torque on a current loop is \( \tau = m B \sin\theta \), where \( m = IA \) is its magnetic moment and \( \theta \) is the angle between \( \vec{m} \) and \( \vec{B} \). The magnetic moment is perpendicular to the plane of the loop, so a field parallel to the plane makes \( \theta = 90^\circ \) and the torque is maximum.
\( \displaystyle A = (1\,\mathrm{cm})^{2} = 1\,\mathrm{cm^2} = 1\times10^{-4}\,\mathrm{m^2} \)
\( \displaystyle \tau = I A B \sin 90^\circ = 10 \times 1\times10^{-4} \times 0.2 \times 1 = 2\times10^{-4}\,\mathrm{N\,m} \)