Question

The magnetic field at the centre of a current carrying circular loop of radius \(R\) is \(16\,\mu\text{T}\). The magnetic field at a distance \(x=\sqrt{3}R\) on its axis from the centre is ____ \(\mu\text{T}\).

• A. 4
• B. 8
• C. \(2\sqrt{2}\)
• D. 2

Answer 1

The Correct Option is B

Concept: The magnetic field on the axis of a circular current-carrying loop at a distance \(x\) from its centre is given by

\[ B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] 

At the centre of the loop (\(x = 0\)), the magnetic field is

\[ B_0 = \frac{\mu_0 I}{2R} \]

Step 1: Given magnetic field at the centre,

\[ B_0 = 16\,\mu\text{T} \]

Step 2: Magnetic field at distance \(x=\sqrt{3}R\) on the axis is

\[ B = B_0 \left( \frac{R^2}{R^2 + x^2} \right)^{3/2} \]

Substituting \(x=\sqrt{3}R\),

\[ B = 16 \left( \frac{R^2}{R^2 + 3R^2} \right)^{3/2} = 16 \left( \frac{1}{4} \right)^{3/2} \]

Step 3: Simplifying,

\[ \left( \frac{1}{4} \right)^{3/2} = \frac{1}{8} \] \[ B = 16 \times \frac{1}{8} = 2 \times 4 = 8\,\mu\text{T} \]