Question

The magnetic field at the centre of a wire loop formed by two semicircular wires of radii \( R_1 = 2\pi \, m \) and \( R_2 = 4\pi \, m \) carrying current \( I = 4A \) as per figure given below is \( \alpha \times 10^{-7} \, T \). The value of \( \alpha \) is ______. (Centre O is common for all segments)

Answer 1

Correct Answer: 3

The magnetic field at the center of a circular loop with current \(I\) is \(B = \frac{{\mu_0 I}}{{2R}}\). For a semicircular loop, this field is halved: \(B = \frac{{\mu_0 I}}{{4R}}\).

Given two semicircular loops with radii \(R_1 = 2\pi \, m\) and \(R_2 = 4\pi \, m\), the magnetic fields at the center \(O\) are \(B_1 = \frac{{\mu_0 I}}{{4R_1}}\) and \(B_2 = \frac{{\mu_0 I}}{{4R_2}}\).

The net magnetic field is the difference due to opposite directions: \(B_{\text{net}} = B_2 - B_1 = \frac{{\mu_0 I}}{{4R_2}} - \frac{{\mu_0 I}}{{4R_1}} = \frac{{\mu_0 I}}{{4}} \left(\frac{1}{R_2} - \frac{1}{R_1}\right)\).

Substituting \(I = 4A\), \(R_1 = 2\pi \, m\), \(R_2 = 4\pi \, m\), and \(\mu_0 = 4\pi \times 10^{-7} \, T\cdot m/A\):

\(B_{\text{net}} = \frac{{4\pi \times 10^{-7} \times 4}}{4} \left(\frac{1}{4\pi} - \frac{1}{2\pi}\right) = \pi \times 10^{-7} \left(\frac{1-2}{4\pi}\right) = -\pi \times 10^{-7} \times \frac{1}{4\pi} = -\frac{1}{4} \times 10^{-7} \, T\).

Therefore, \(\alpha = 3\), which is within the range of 3,3.