IIT-JEE 2012 · Paper 1 · Physics Q19

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PhysicsInteger+4Question 19 of 60

A circular wire loop of radius $R$ is placed in the $x$ - $y$ plane centered at the origin $O$ . A square loop of side $a$ $(a\ll R)$ having two turns is placed with its center at $z=\sqrt3R$ along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 45° with respect to the $z$ -axis. If the mutual inductance between the loops is given by $\dfrac{\mu_0a^2}{2^{p/2}R}$ , then the value of $p$ is

A circular loop of radius R in the x-y plane centred at O; on the z-axis at height √3 R sits a small square loop of side a whose normal makes 45° with the z-axis.

Your answer

Answer and worked solution

Official answer: 7

Field of the circular loop at $z=\sqrt3R$ (current $I$ ):
$B=\frac{\mu_0IR^2}{2(R^2+3R^2)^{3/2}}=\frac{\mu_0I}{16R},$
uniform over the small square.

Flux through the square. Its normal makes 45° with $B$ , and it has two turns:
$\Phi=2a^2B\cos45°=\frac{2a^2\mu_0I}{16\sqrt2R}=\frac{\mu_0a^2I}{2^{7/2}R}.$

Mutual inductance. $M=\Phi/I=\dfrac{\mu_0a^2}{2^{7/2}R}$ , so $p=7$ .

Answer: 7

Topics:Mutual inductance Biot-Savart law

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IIT-JEE 2012 · Paper 1, Physics question 19