Question

A circular coil of radius R carries current such that magnetic field at its centre is 16\(\mu\)T. Find the magnetic field on the axis at a distance of \(\sqrt{3}R\) from the centre of coil.

• A. 2\(\mu\)T
• B. 4\(\mu\)T
• C. 3\(\mu\)T
• D. 5\(\mu\)T
• E. None of these

Hint:

For distance \(x = R\sqrt{3}\), the denominator factor is always 8. For \(x = R\), it's \(2\sqrt{2}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The magnetic field of a current loop decreases as we move along its central axis away from the center.
Step 2: Key Formula or Approach:
Field at center: \(B_c = \frac{\mu_0 I}{2R}\).
Field on axis at distance \(x\): \(B_a = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}\).
Ratio: \(\frac{B_a}{B_c} = \left( \frac{R^2}{R^2 + x^2} \right)^{3/2}\).
Step 3: Detailed Explanation:
Given \(x = \sqrt{3}R\).
\[ \frac{B_a}{B_c} = \left( \frac{R^2}{R^2 + (\sqrt{3}R)^2} \right)^{3/2} = \left( \frac{R^2}{R^2 + 3R^2} \right)^{3/2} \]
\[ \frac{B_a}{B_c} = \left( \frac{R^2}{4R^2} \right)^{3/2} = \left( \frac{1}{4} \right)^{3/2} = \frac{1}{(4^{1/2})^3} = \frac{1}{2^3} = \frac{1}{8} \]
Given \(B_c = 16 \text{ \(\mu\)T}\):
\[ B_a = \frac{B_c}{8} = \frac{16}{8} = 2 \text{ \(\mu\)T} \]
Step 4: Final Answer:
The magnetic field at the given axial point is 2 \(\mu\)T.