When a current carrying coil is placed in a uniform external magnetic…

The correct answer is Option (2) → (B), (D) and (E) only

For a current carrying coil of magnetic moment \( \vec{m} \) in a uniform field \( \vec{B} \), the potential energy is \( U = -\vec{m}\cdot\vec{B} = -mB\cos\theta \) and the torque is \( \vec{\tau} = \vec{m}\times\vec{B} \), with magnitude \( mB\sin\theta \). Antiparallel means \( \theta = 180^\circ \).

    \( \displaystyle U = -mB\cos180^\circ = +mB \qquad \tau = mB\sin180^\circ = 0 \)

(A) Its potential energy is minimum. — Incorrect. The minimum is \( -mB \), at \( \theta = 0^\circ \) (parallel); at \( 180^\circ \) the energy is maximum.

(B) Its potential energy is maximum. — Correct. \( U = +mB \) is the largest value it can take.

(C) Torque on it is maximum. — Incorrect. The torque is maximum at \( \theta = 90^\circ \); at \( 180^\circ \) it is zero.

(D) Torque on it is zero. — Correct. \( \sin180^\circ = 0 \).

(E) Coil is in unstable equilibrium. — Correct. The net torque is zero, but a small rotation produces a torque that turns the coil further away, because the energy is at a maximum; this is unstable equilibrium.

So the correct statements are (B), (D) and (E), which is Option (2).

NCERT (2026-27):
The Dipole in a Uniform Magnetic Field – “… The torque on the needle is [see Eq. (4.23)],

    \( \displaystyle \boldsymbol{\tau} = \mathbf{m} \times \mathbf{B} \qquad (5.2) \)

In magnitude

    \( \displaystyle \tau = mB\sin\theta \)”

The Dipole in a Uniform Magnetic Field – “… Equation (5.3) shows that potential energy is minimum (= –mB) at \( \theta = 0^\circ \) (most stable position) and maximum (= +mB) at \( \theta = 180^\circ \) (most unstable position).”