JEE Main 2016 · 3 Apr · Shift 1 · Physics Q15

← JEE Main 2016 · 3 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 15 of 90

Two identical wires A and B, each of length ' $l$ ', carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ' $a$ '. If $B_A$ and $B_B$ are the values of magnetic field at the centres of the circle and square respectively, then the ratio $\dfrac{B_A}{B_B}$ is :

\dfrac{\pi^2}8

\dfrac{\pi^2}{16\sqrt2}

\dfrac{\pi^2}{16}

\dfrac{\pi^2}{8\sqrt2}

Answer and worked solution

Official answer: D

Circle: $R=\frac l{2\pi}$ , so $B_A=\frac{\mu_0I}{2R}=\frac{\pi\mu_0I}l$ . Square of side $a=\frac l4$ : each side gives $\frac{\mu_0I}{4\pi(a/2)}\cdot2\sin45^\circ$ , so
$B_B=4\cdot\frac{\sqrt2\mu_0I}{2\pi a}=\frac{8\sqrt2\mu_0I}{\pi l}.$
The ratio is $\dfrac{B_A}{B_B}=\dfrac{\pi^2}{8\sqrt2}$ .

Topics:Biot-Savart law

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JEE Main 2016 · 3 Apr · Shift 1, Physics question 15