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 Question:
We must calculate the magnetic field $B$ produced by an $n$-turn circular coil on its central axis at a specific distance $x$.

Step 2: Key Formula or Approach:
The magnetic field at an axial distance $x$ from the center of a circular coil of radius $r$ carrying current $I$ with $n$ turns is:
$$B = \frac{\mu_0 n I r^2}{2(r^2 + x^2)^{3/2}}$$

Step 3: Detailed Explanation:
We are given the distance $x = 2\sqrt{2} r$.
First, let's calculate the denominator term $(r^2 + x^2)$:
$$x^2 = (2\sqrt{2} r)^2 = 8r^2$$
$$r^2 + x^2 = r^2 + 8r^2 = 9r^2$$
Now, raise this to the power of $3/2$:
$$(9r^2)^{3/2} = (\sqrt{9r^2})^3 = (3r)^3 = 27r^3$$
Substitute this back into the main magnetic field formula:
$$B = \frac{\mu_0 n I r^2}{2(27r^3)}$$
Simplify the fraction:
$$B = \frac{\mu_0 n I}{54r}$$

Step 4: Final Answer:
The magnetic field is $\frac{\mu_0 n I}{54r}$, matching option (c).