Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) Magnetised needle at \(60^\circ\) to a uniform field | (I) Moves from the weaker to the stronger part of the field |
| (B) Magnetised needle with its moment along a uniform field | (II) Feels a torque but no net force |
| (C) Paramagnetic bar in a non-uniform field | (III) Moves from the stronger to the weaker part of the field |
| (D) Diamagnetic bar in a non-uniform field | (IV) Feels no force and no torque, and is in stable equilibrium |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(A) Magnetised needle at \( 60^\circ \) to a uniform field — This matches (II). The net force in a uniform field is zero, but the torque \( mB\sin60^\circ \) is not zero.
(B) Magnetised needle with its moment along a uniform field — This matches (IV). At \( \theta=0^\circ \) the torque \( mB\sin\theta \) is zero, and the energy \( -mB \) is the lowest, so this is the most stable position.
(C) Paramagnetic bar in a non-uniform field — This matches (I). It is weakly attracted, so it moves towards the stronger field.
(D) Diamagnetic bar in a non-uniform field — This matches (III). It is weakly repelled, so it moves towards the weaker field.
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, …”
The dipole in a uniform magnetic field – “… potential energy is minimum (= –mB) at θ = 0° (most stable position) and maximum (= +mB) at θ = 180° (most unstable position).”
Paramagnetism – “… When placed in a non-uniform magnetic field, the bar will tend to move from weak field to strong.”
Diamagnetism – “… When placed in a non-uniform magnetic field, the bar will tend to move from high to low field.”