Question

The magnetic field due to a current carrying circular loop of radius \(6\ \text{cm}\) at a point on the axis at a distance of \(8\ \text{cm}\) from its centre is \(27\ \mu\text{T}\). The magnetic field at the centre of the current carrying loop is

• A. \(75\ \mu\text{T}\)
• B. \(125\ \mu\text{T}\)
• C. \(150\ \mu\text{T}\)
• D. \(250\ \mu\text{T}\)

Hint:

For a circular current loop, \[ B_{\text{centre}}=\frac{\mu_0 I}{2R}, \] and at a point on the axis, \[ B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}. \] Taking the ratio of these two formulas often eliminates the current \(I\) and simplifies the calculation.

Answer 1

The Correct Option is B

Step 1: Relate on-axis field to centre field.
Field at center: \( B_0 = \frac{\mu_0 I}{2R} \). Field on axis at distance \( x \): \( B_x = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}} = B_0\cdot\frac{R^3}{(R^2+x^2)^{3/2}} \).

Step 2: Find \( B_0 \).
\( R = 6\text{ cm},\; x = 8\text{ cm} \Rightarrow (R^2+x^2)^{3/2} = (36+64)^{3/2} = 100^{3/2} = 1000 \).
\( \frac{B_x}{B_0} = \frac{R^3}{1000} = \frac{216}{1000} \Rightarrow B_0 = \frac{27\times10^{-6}\times1000}{216} = 125\,\mu\text{T} \)

\[ \boxed{B_0 = 125\,\mu\text{T}} \]