IIT-JEE 2012 · Paper 2 · Physics Q1

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PhysicsSingle correct+3 / -1Question 1 of 60

A loop carrying current $I$ lies in the $x$ - $y$ plane as shown in the figure. The unit vector $\hat k$ is coming out of the plane of the paper. The magnetic moment of the current loop is

A current loop made of a square of side a centred at the origin with a semicircle of diameter a bulging outward on each of its four sides (a four-leaf shape); the current I flows counter-clockwise.

a^2I\hat k

\left(\dfrac\pi2+1\right)a^2I\hat k

-\left(\dfrac\pi2+1\right)a^2I\hat k

(2\pi+1)a^2I\hat k

Answer and worked solution

Official answer: B

Area. The loop is a square of side $a$ with a semicircle of diameter $a$ on each side:
$A=a^2+4\cdot\frac12\pi\left(\frac a2\right)^2=a^2+\frac{\pi a^2}2=\left(\frac\pi2+1\right)a^2.$

Direction. The current is counter-clockwise as seen from the front, so the moment points out of the paper, along $+\hat k$ .

Moment. $\vec m=IA\hat k=\left(\frac\pi2+1\right)a^2I\hat k$ .

(B)

Topics:Torque on a current loop

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IIT-JEE 2012 · Paper 2, Physics question 1