Physics Circle home Student Login

Home›Chapters›Moving Charges and Magnetism PYQs›Biot-Savart Law & Circular Coils›Question 3

MHT CET 2026 (16 Apr, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2026 (16 Apr, Shift 2)

Q3MHT CET 2026 · 16 APR · SHIFT 2Easy

Biot-Savart Law & Circular Coils

A circular loop of radius R is carrying current I. The ratio of magnetic field at the center of circular loop and at a distance R from the center of the loop on its axis is

Show answer and solution

Answer: (c)

  1. Bc=μ0I2RB_c = \frac{\mu_0 I}{2R}, Bx=μ0IR22(R2+x2)3/2B_x = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}
  2. BcBx=(R2+x2)3/2R3\frac{B_c}{B_x} = \frac{(R^2+x^2)^{3/2}}{R^3}; with x=Rx = R: (2R2)3/2R3=22=8\frac{(2R^2)^{3/2}}{R^3} = 2\sqrt{2} = \sqrt{8}
  3. So the answer is (c) 8:1\sqrt{8}:1.
Report a mistake

More questions on Biot-Savart Law & Circular Coils