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

To solve this problem, we need to find the magnetic field at a distance \(x=\sqrt{3}R\) on the axis of a current-carrying circular loop, given the magnetic field at the center of the loop. Let's break down the problem step-by-step:

1. The magnetic field at the center of a circular loop carrying current \(I\) is given by: \(B_{\text{center}} = \frac{\mu_0 I}{2R}\), where \(\mu_0\) is the permeability of free space. 
2. The magnetic field at a point on the axis of a circular loop at a distance \(x\) from its center is given by: \(B_{\text{axis}} = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}\).
3. According to the problem, the magnetic field at the center is \(16\,\mu\text{T}\). Therefore: 
   \(16 \times 10^{-6} = \frac{\mu_0 I}{2R}\) 
   Solving for \(I\): 
   \(I = \frac{\left(16 \times 10^{-6}\right) \cdot 2R}{\mu_0}\).
4. We need to find \(B_{\text{axis}}\) at \(x = \sqrt{3}R\). Substituting \(x = \sqrt{3}R\) in the \(B_{\text{axis}}\) formula: 
   \(B_{\text{axis}} = \frac{\mu_0 I R^2}{2(R^2+(\sqrt{3}R)^2)^{3/2}}\) 
   Simplifying the expression: 
   \(B_{\text{axis}} = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}}\) 
   \(B_{\text{axis}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}}\) 
   \(B_{\text{axis}} = \frac{\mu_0 I R^2}{2(8R^3)}\) 
   \(B_{\text{axis}} = \frac{\mu_0 I}{16R}\)
5. Substitute the expression for \(I\) found earlier: 
   \(B_{\text{axis}} = \frac{\mu_0 \cdot \left(\frac{16 \times 10^{-6} \cdot 2R}{\mu_0}\right)}{16R}\) 
   \(B_{\text{axis}} = \frac{32 \times 10^{-6}}{16}\) 
   \(B_{\text{axis}} = 2 \times 10^{-6} \cdot 8\) 
   \(B_{\text{axis}} = 8 \mu\text{T}\).

Thus, the magnetic field at a distance \(x = \sqrt{3}R\) on its axis is 8 \(\mu\text{T}\). Therefore, the correct answer is option 8.