Consider two thin identical conducting wires covered with very thin… | JEE Main 2014 (12 Apr, Shift 1)

Q95JEE Main 2014 · 12 APR · SHIFT 1Medium

Biot-Savart Law & Circular Coils

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 B1, at its center 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 B2 at the centre of the loops when current I/3 passes through it. The ratio B1: B2 is:

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Answer: (b)

Loop: $B_1 = \frac{\mu_0 I}{2R}$ .
Coil of 3 turns from the same length: radius $R/3$ , current $I/3$ : $B_2 = \frac{\mu_0 \cdot 3 \cdot (I/3)}{2(R/3)} = \frac{3\mu_0 I}{2R}$ .
$B_1 : B_2 = 1 : 3$
So the answer is (b) 1: 3.

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Biot-Savart Law & Circular Coils question from JEE Main 2014 (12 Apr, Shift 1)

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