Question

Figure shows a current carrying square loop ABCD of edge length is $ a $ lying in a plane. If the resistance of the ABC part is $ r $ and that of the ADC part is $ 2r $, then the magnitude of the resultant magnetic field at the center of the square loop is:

• A. \( \frac{\sqrt{2}\mu_0 I}{3 \pi a} \)
• B. \( \frac{\mu_0 I}{2 \pi a} \)
• C. \( \frac{2 \mu_0 I}{3 \pi a} \)
• D. \( \frac{3 \pi \mu_0 I}{\sqrt{2}} \)

Hint:

When dealing with complex current-carrying loops, always break the problem into simpler segments, calculate their magnetic fields individually, and then combine them using superposition.

Answer 1

The Correct Option is A

From the provided diagram, the following data is established:

\( R_{ABC} = r \), and \( R_{ADC} = 2r \)

The currents are defined as follows:

\( i_1 = \frac{2I}{3} \), and \( i_2 = \frac{I}{3} \)

The magnetic field at the center, \( B_{\text{centre}} \), is computed using the formula:

\( B_{\text{centre}} = \frac{2\mu_0 I \sqrt{2}}{4 \pi \left( \frac{a}{2} \right)} \left[ \frac{2I}{3} - \frac{I}{3} \right] = \sqrt{2} \frac{\mu_0 I}{3\pi a} \)