A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below: If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium? (a) (B) and (D), respectively (b) (B) and (C), respectively (c) (A) and (B), respectively (d) (A) and (C), respectively

Answer: (a). Stable equilibrium: m parallel to B ( θ = 0, U = -mB minimum). Unstable equilibrium: m antiparallel to B ( θ = 180°, U = +mB maximum). In (A) and (C) the loop is in a vertical plane, so m ⊥ B and there is a torque. In (B) the current gives m along +z and in (D) along -z. So the answer is (a) (B) and (D), respectively.

A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12… | JEE Main 2015

Q81JEE Main 2015Medium

Current Loop as a Magnetic Dipole

A rectangular loop of sides $10\text{ cm}$ and $5\text{ cm}$ carrying a current $I$ of $12\text{ A}$ is placed in different orientations as shown in the figures below:

If there is a uniform magnetic field of $0.3\text{ T}$ in the positive $z$ direction, in which orientations the loop would be in $(i)$ stable equilibrium and $(ii)$ unstable equilibrium?

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Answer: (a)

Stable equilibrium: $\vec m$ parallel to $\vec B$ ( $\theta = 0$ , $U = -mB$ minimum).
Unstable equilibrium: $\vec m$ antiparallel to $\vec B$ ( $\theta = 180^\circ$ , $U = +mB$ maximum).
In (A) and (C) the loop is in a vertical plane, so $\vec m \perp \vec B$ and there is a torque.
In (B) the current gives $\vec m$ along $+z$ and in (D) along $-z$ .
So the answer is (a) (B) and (D), respectively.

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