Consider the following statements about a short bar magnet of magnetic moment \(\vec{m}\):
(A) In a uniform magnetic field, the net force on it is zero.
(B) In a uniform magnetic field \(\vec{B}\), its potential energy \(-\vec{m}\cdot\vec{B}\) is zero when \(\vec{m}\) is perpendicular to \(\vec{B}\).
(C) Its magnetic field at a far point on its axis varies as \(1/r^{2}\), where \(r\) is the distance from its centre.
(D) Its magnetic field at a far point on its equatorial line is in the direction of \(\vec{m}\).
Which of the above statements are incorrect?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C) and (D) only
A uniform field turns a bar magnet but does not move it as a whole. Far from a short magnet, its field falls off as \( 1/r^{3} \), both on the axis and on the equatorial line.
(A) In a uniform magnetic field, the net force on it is zero. — Correct statement. The magnet feels only a torque \( \vec{m}\times\vec{B} \) in a uniform field.
(B) In a uniform magnetic field \(\vec{B}\), its potential energy \(-\vec{m}\cdot\vec{B}\) is zero when \(\vec{m}\) is perpendicular to \(\vec{B}\). — Correct statement. Here \( U=-mB\cos\theta \) and \( \cos 90^\circ \) is zero. This orientation is chosen as the zero of energy.
(C) Its magnetic field at a far point on its axis varies as \(1/r^{2}\), where \(r\) is the distance from its centre. — Incorrect statement. The far axial field is \( B_A=\mu_0 2m/4\pi r^{3} \). It varies as \( 1/r^{3} \), like the axial field of an electric dipole.
(D) Its magnetic field at a far point on its equatorial line is in the direction of \(\vec{m}\). — Incorrect statement. The equatorial field \( B_E=-\mu_0 m/4\pi r^{3} \) carries a minus sign. It points opposite to \( \vec{m} \).
NCERT (2026-27):
Summary – “2. When a bar magnet of dipole moment m is placed in a uniform magnetic field B, (a) the force on it is zero, (b) the torque on it is m × B, (c) its potential energy is –m.B, where we choose the zero of energy at the orientation when m is perpendicular to B.”
Summary – “3. Consider a bar magnet of size l and magnetic moment m, … where r >>l, the magnetic field B due to this bar”
\( \displaystyle B=\frac{\mu_0 m}{2\pi r^{3}} \) \( \text{(along axis)} \) \( \displaystyle =-\frac{\mu_0 m}{4\pi r^{3}} \) \( \text{(along equator)} \)