Question

The ratio of magnetic field at the centre of the current carrying circular loop and its magnetic moment is '\(x\)'. When both the current and radius are tripled, then the ratio will be

• A. \(\frac{x}{27}\)
• B. \(\frac{x}{9}\)
• C. \(\frac{x}{3}\)
• D. \(3x\)

Hint:

Write both quantities in terms of I and R; the current cancels in the ratio.

Answer 1

The Correct Option is A

Step 1: New values:
$B' = \frac{\mu_0(3I)}{2(3R)} = B$. So the field does not change.

Step 2: Magnetic moment:
$m' = (3I)\pi(3R)^2 = 27m$. So the ratio $\frac{B'}{m'} = \frac{B}{27m} = \frac{x}{27}$.

Final Answer:
The ratio becomes $\frac{x}{27}$, option (A). \[ \boxed{\frac{x}{27}} \]