Question

A circular coil carrying current 'I' has a radius 'r' and 'n' turns. The magnetic field along the axis of a coil at a distance '2√2 r' from its centre is ______.

• A. $\frac{\mu_0 n I}{9r}$
• B. $\frac{\mu_0 n I}{18r}$
• C. $\frac{\mu_0 n I}{54r}$
• D. $\frac{\mu_0 n I}{27r}$

Hint:

To avoid algebra mistakes with the $3/2$ power, always take the square root of the bracket first (power of $1/2$), and then cube the result. It keeps the numbers much smaller and easier to manage mentally!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The magnetic field $B$ on the axis of a circular coil of $n$ turns at distance $x$ is $B = \frac{\mu_0 n I r^2}{2(r^2 + x^2)^{3/2}}$.

Step 2: Formula Application:
Substitute $x = 2\sqrt{2} r$. $x^2 = (2\sqrt{2} r)^2 = 8r^2$. Denominator: $(r^2 + 8r^2)^{3/2} = (9r^2)^{3/2} = (3r)^3 = 27r^3$.

Step 3: Explanation:
$B = \frac{\mu_0 n I r^2}{2(27r^3)} = \frac{\mu_0 n I}{54r}$.

Step 4: Final Answer:
The magnetic field is $\frac{\mu_0 n I}{54r}$.