JEE Main 2015 · 4 Apr · Shift 1 · Physics Q21

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PhysicsSingle correct+4 / -1Question 21 of 90

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 ?

Four orientations of the loop with B along +z: (a) in the y-z plane, (b) in the x-y plane, (c) in the x-z plane, (d) in the x-y plane with the current circulating in the opposite sense to (b).

(a) and (b), respectively

(a) and (c), respectively

(b) and (d), respectively

(b) and (c), respectively

Answer and worked solution

Official answer: C

Energy $U=-\vec m\cdot\vec B$ .

In (b) the loop lies in the $x$ - $y$ plane with $\vec m$ along $+z$ , parallel to $\vec B$ : minimum energy, stable.
In (d) the current circulates the other way, so $\vec m$ is antiparallel to $\vec B$ : maximum energy, unstable.
In (a) and (c) $\vec m$ is perpendicular to $\vec B$ , and there is a torque.

So (b) and (d).

Topics:Torque on a current loop

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 21