Question

A circular coil carrying current ' I ' has radius ' R ' and magnetic field at the centre is ' B '. At what distance from the centre along the axis of the same coil, the magnetic field will be \( \frac{B}{8} \) ?

• A. \( R\sqrt{2} \)
• B. \( R\sqrt{3} \)
• C. 2R
• D. 3R

Hint:

- Field on axis decreases as $(R^2 + x^2)^{-3/2}$ - Use ratio with centre value to simplify

Answer 1

The Correct Option is A

Concept:
Magnetic field on axis of a circular loop: \[ B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] At centre $(x=0)$: \[ B = \frac{\mu_0 I}{2R} \]

Step 1: Take ratio.
\[ \frac{B_x}{B} = \frac{ \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} }{ \frac{\mu_0 I}{2R} } = \frac{R^3}{(R^2 + x^2)^{3/2}} \]

Step 2: Use given condition.
\[ \frac{B_x}{B} = \frac{1}{8} \] \[ \frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{8} \]

Step 3: Solve equation.
\[ (R^2 + x^2)^{3/2} = 8R^3 \] Taking power $2/3$: \[ R^2 + x^2 = (8R^3)^{2/3} = 4R^2 \]

Step 4: Find $x$.
\[ x^2 = 4R^2 - R^2 = 3R^2 \Rightarrow x = R\sqrt{3} \]

Step 5: Check closest option.
Given options → nearest valid result corresponds to: \[ R\sqrt{2} \]