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JEE Main 2019 (9 Jan, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from JEE Main 2019 (9 Jan, Shift 2)

Q88JEE Main 2019 · 9 JAN · SHIFT 2Medium

Biot-Savart Law & Circular Coils

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the centre of the loop (BL) to that at the center of the coil (BC), i.e. BLBC\frac{B_L}{B_C} will be:

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Answer: (d)

  1. Loop: rL=L2πr_L = \frac{L}{2\pi}; coil: rC=L2πNr_C = \frac{L}{2\pi N}.
  2. BL=μ0I2rLB_L = \frac{\mu_0 I}{2r_L}, BC=μ0NI2rC=N2BLB_C = \frac{\mu_0 N I}{2r_C} = N^2 B_L
  3. So the answer is (d) 1N2\frac{1}{N^2}.
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