Question

Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is _____________

• A. 0.1 m
• B. 0.15 m
• C. 0.2 m
• D. 1.0 m

Hint:

The magnetic field is maximum at the center ($x=0$) and decreases as you move away along the axis. The $3/2$ power in the denominator is key to these ratio problems.

Answer 1

The Correct Option is A

Step 1: $B_{axis} = \frac{\mu_0 i R^2}{2(R^2 + x^2)^{3/2}}$. So $B \propto (R^2 + x^2)^{-3/2}$.
Step 2: $\frac{B_1}{B_2} = \left( \frac{R^2 + x_2^2}{R^2 + x_1^2} \right)^{3/2} = 8/1$.
Step 3: Take cube root of both sides: $\sqrt{\frac{R^2 + x_2^2}{R^2 + x_1^2}} = 2$.
Step 4: Square both sides: $\frac{R^2 + 0.2^2}{R^2 + 0.05^2} = 4$.
Step 5: $R^2 + 0.04 = 4R^2 + 4(0.0025) = 4R^2 + 0.01$.
Step 6: $3R^2 = 0.03 \Rightarrow R^2 = 0.01 \Rightarrow R = 0.1$ m.