A circular loop of area \( 1\,\mathrm{cm^2} \), carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → zero
A current loop has a magnetic moment \( \vec{m} = I\vec{A} \), directed along the normal to its plane. In a uniform field the torque on the loop is
\( \displaystyle \tau = m B \sin\theta \)
where \( \theta \) is the angle between \( \vec{m} \) and \( \vec{B} \). The field is perpendicular to the plane of the loop, so it lies along the normal, parallel to \( \vec{m} \):
\( \displaystyle \theta = 0^\circ \;\Rightarrow\; \tau = m B \sin 0^\circ = 0 \)
NCERT (2026-27):
Torque on Current Loop, Magnetic Dipole – “… We define the magnetic moment of the current loop as,
\( \displaystyle \mathbf{m} = I\mathbf{A} \qquad (4.22) \)
where the direction of the area vector A is given by the right-hand thumb rule and is directed into the plane of the paper in Fig. 4.18. … From Eq. (4.23), we see that the torque τ vanishes when m is either parallel or antiparallel to the magnetic field B.”