Question

A circular coil having '\(n\)' turns, each of radius '\(r\)' carries a current '\(I\)'. The magnetic field at a distance \(x=\sqrt{3}r\) from its center, along the axis of the coil, is \([\mu _{0}\) = permeability of free space\(]\)

• A. \(\frac{\mu _{0}nI}{4r}\)
• B. \(\frac{\mu _{0}nI}{8r}\)
• C. \(\frac{\mu _{0}nI}{16r}\)
• D. \(\frac{\mu _{0}nI}{32r}\)

Hint:

Use the axial field formula of a coil and put \(x=\sqrt{3}\,r\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a coil of \(n\) turns the axial field is \(B=\dfrac{\mu_{0}nIr^{2}}{2(r^{2}+x^{2})^{3/2}}\).

Step 2: Substitute \(x=\sqrt{3}r\)
\(r^{2}+x^{2}=4r^{2}\), so \((4r^{2})^{3/2}=8r^{3}\).

Step 3: Calculation
\(B=\dfrac{\mu_{0}nIr^{2}}{2\times8r^{3}}=\dfrac{\mu_{0}nI}{16r}\)

Final Answer:
The magnetic field at that point is μ0 n I/(16 r). \[ \boxed{\text{(C) }\dfrac{\mu_{0}nI}{16r}} \]