Question

A circular coil having 'n' turns, each of radius 'r' carries a current 'I'. The magnetic field at a distance \(x=\sqrt{8}\ r\) from its centre along axis of the coil is \((\mu _{0}=\) permeability of free space)

• A. \(\frac{\mu _{0}nI}{54\ r}\)
• B. \(\frac{\mu _{0}nI}{27\ r}\)
• C. \(\frac{\mu _{0}nI}{64\ r}\)
• D. \(\frac{\mu _{0}nI}{36\ r}\)

Hint:

Use the axial field formula and substitute x squared equal to 8 r squared.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
On the axis of a coil of \(n\) turns: \(B=\dfrac{\mu_0nIr^{2}}{2(r^{2}+x^{2})^{3/2}}\).

Step 2: Substitute
\(x^{2}=8r^{2}\Rightarrow r^{2}+x^{2}=9r^{2}\), and \((9r^{2})^{3/2}=27r^{3}\).

Step 3: Calculation
\(B=\dfrac{\mu_0nIr^{2}}{2\times27r^{3}}=\dfrac{\mu_0nI}{54r}\)

Final Answer:
The field is mu-naught n I over 54 r. \[ \boxed{\text{(A) }\dfrac{\mu_{0}nI}{54\ r}} \]