JEE Main 2014 · 12 Apr · Shift 1 · Physics Q20

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PhysicsSingle correct+4 / -1Question 20 of 90

Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field $B_1$ , at its centre when a current I passes through it. The second wire is bent into a coil with three identical loops adjacent to each other and produces magnetic field $B_2$ at the centre of the loops when current I/3 passes through it. The ratio $B_1:B_2$ is :

1 : 1

1 : 3

1 : 9

9 : 1

Answer and worked solution

Official answer: B

Same wire length: one loop of radius $R$ , or three loops of radius $\frac R3$ .
$B_1=\frac{\mu_0I}{2R},\qquad B_2=3\cdot\frac{\mu_0(I/3)}{2(R/3)}=\frac{3\mu_0I}{2R}.$
So $B_1:B_2=1:3$ .

Topics:Biot-Savart law

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 20