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 semicircular wire carrying current \(I\) and having radius \(R\) is given by:

\(B = \frac{\mu_0 I}{4R}.\)

For the semicircular wires of radii \(R_1\) and \(R_2\):

\(B_{R_1} = \frac{\mu_0 I}{4R_1}, \quad B_{R_2} = \frac{\mu_0 I}{4R_2}.\)

The net magnetic field at the center \(O\) is the sum of the fields due to both semicircular wires:

\(B = B_{R_1} + B_{R_2} = \frac{\mu_0 I}{4R_1} + \frac{\mu_0 I}{4R_2}.\)

Substituting the given values:

\(B = \frac{4\pi \times 10^{-7} \cdot 4}{4 \cdot 2} + \frac{4\pi \times 10^{-7} \cdot 4}{4 \cdot 4}.\)

Simplify:

\(B = \pi \times 10^{-7} + \frac{\pi \times 10^{-7}}{2} = 2\pi \times 10^{-7} + \pi \times 10^{-7} = 3\pi \times 10^{-7} \, \text{T}.\)

Therefore:

\(\alpha = 3.\)
The Correct answer is: 3