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 carrying current \(I\) is given by:

\(B = \frac{{\mu_0 I}}{{2R}}\)

For a semicircular loop, the magnetic field is half of that:

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

Given two semicircular loops with radii \(R_1 = 2\pi \, m\) and \(R_2 = 4\pi \, m\):

The magnetic field at the center \(O\) due to each is:

\(B_1 = \frac{{\mu_0 I}}{{4R_1}}\)

\(B_2 = \frac{{\mu_0 I}}{{4R_2}}\)

The net magnetic field at the center is the difference due to opposite directions:

\(B_{\text{net}} = B_2 - B_1\)

\(B_{\text{net}} = \frac{{\mu_0 I}}{{4R_2}} - \frac{{\mu_0 I}}{{4R_1}}\)

\(B_{\text{net}} = \frac{{\mu_0 I}}{{4}} \left(\frac{1}{R_2} - \frac{1}{R_1}\right)\)

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

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

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

\(B_{\text{net}} = -\pi \times 10^{-7} \times \frac{1}{4\pi}\)

\(B_{\text{net}} = -\frac{1}{4} \times 10^{-7} \, T\)

Thus, \(\alpha = 3\). Hence, the value of \(\alpha\) fits within the given range of 3,3.

Answer 2

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