Question

A circular coil carrying current ' T ' has a radius ' \(r\) ' and ' \(n\) ' turns. The magnetic field along the axis of a coil at a distance ' \(2 \sqrt{2} r\) ', from its centre is ( \(\mu_0=\) permeability of free space, \(n\) is very small)

• A. \(\frac{\mu _{0}\mathrm{nI}}{9\mathrm{r}}\)
• B. \(\frac{\mu _{0}nI}{18r}\)
• C. \(\frac{\mu _{0}\mathrm{nI}}{54\mathrm{r}}\)
• D. \(\frac{\mu _{0}\mathrm{nI}}{27\mathrm{r}}\)

Hint:

Use the on-axis field formula of a circular coil and put the axial distance equal to 2 root 2 times the radius.

Answer 1

The Correct Option is C

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

Step 2: Substitute x
With \(x=2\sqrt2\,r\): \(x^2=8r^2\), so \(r^2+x^2=9r^2\) and \((9r^2)^{3/2}=27r^3\).

Step 3: Calculation
\(B=\dfrac{\mu_0nIr^2}{2\times27r^3}=\dfrac{\mu_0nI}{54r}\). This matches option (C).

Final Answer:
The field is mu-zero n I divided by 54 r. \[ \boxed{\text{(C) }\dfrac{\mu_{0}nI}{54r}} \]