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

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