Question

A circular coil of radius \(R\) carries a current \(I\). The magnetic field at its centre is \(B\). At what distance from the centre on the axis of the coil will the magnetic field be \(B/8\)?

• A. \(R\)
• B. \(2R\)
• C. \(\sqrt{3}R\)
• D. \(4R\)

Hint:

Divide the axial-field expression by the field at the centre first. Work with the dimensionless variable $x/R$ so all constants cancel immediately.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are comparing the magnetic field strength at the center of a circular loop with the field strength at a point along its axial line.
Step 2: Key Formula or Approach:
Magnetic field at the center (\( B_c \)):
\[ B_c = \frac{\mu_0 I}{2R} \]
Magnetic field on the axis at distance \( x \) (\( B_x \)):
\[ B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \]
Step 3: Detailed Explanation:
Given that \( B_x = \frac{B_c}{8} \).
Substitute the formulas:
\[ \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{1}{8} \left( \frac{\mu_0 I}{2R} \right) \]
Cancel common terms (\( \frac{\mu_0 I}{2} \)) from both sides:
\[ \frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{8R} \]
Rearranging the equation:
\[ 8R^3 = (R^2 + x^2)^{3/2} \]
Taking the cube root (raising to power \( 1/3 \)) on both sides:
\[ (8R^3)^{1/3} = ((R^2 + x^2)^{3/2})^{1/3} \]
\[ 2R = (R^2 + x^2)^{1/2} \]
Squaring both sides:
\[ 4R^2 = R^2 + x^2 \]
\[ 3R^2 = x^2 \]
\[ x = \sqrt{3}R \]
Step 4: Final Answer:
The magnetic field becomes \( B/8 \) at a distance of \( \sqrt{3}R \) from the center.