A bar magnet of magnetic moment \(m\) is placed in a uniform magnetic field \(B\). The zero of potential energy is taken when the magnet is perpendicular to the field.
Match List-I with List-II.
| List-I (Angle between \(\vec m\) and \(\vec B\)) | List-II (Potential energy) |
|---|---|
| (A) \(0^\circ\) | (I) \(+\frac{1}{2}mB\) |
| (B) \(60^\circ\) | (II) \(-mB\) |
| (C) \(120^\circ\) | (III) \(+mB\) |
| (D) \(180^\circ\) | (IV) \(-\frac{1}{2}mB\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
The potential energy of a magnet in a uniform field is \( U=-mB\cos\theta \). Here \( \theta \) is the angle between the magnetic moment and the field. The sign of \( \cos\theta \) decides the sign of \( U \).
(A) \( 0^\circ \) — This matches (II) \( -mB \). Since \( \cos0^\circ=1 \), \( U=-mB \). This is the most stable position.
(B) \( 60^\circ \) — This matches (IV) \( -\frac{1}{2}mB \). Since \( \cos60^\circ=\frac{1}{2} \), \( U=-\frac{1}{2}mB \).
(C) \( 120^\circ \) — This matches (I) \( +\frac{1}{2}mB \). Since \( \cos120^\circ=-\frac{1}{2} \), \( U=+\frac{1}{2}mB \).
(D) \( 180^\circ \) — This matches (III) \( +mB \). Since \( \cos180^\circ=-1 \), \( U=+mB \). This is the most unstable position.
NCERT (2026-27):
The dipole in a uniform magnetic field – “The magnetic potential energy Um is given by”
\( U_m=\int\tau(\theta)\,d\theta=\int mB\sin\theta\,d\theta= \)\( -mB\cos\theta \)\( =-\mathbf{m}\cdot\mathbf{B} \qquad (5.3) \)
The dipole in a uniform magnetic field – “… Equation (5.3) shows that potential energy is minimum (= –mB) at θ = 0° (most stable position) and maximum (= +mB) at θ = 180° (most unstable position).”