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

Given the diagram, we have the following information:

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

Also, the currents \( i_1 \) and \( i_2 \) are given by:

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

The magnetic field at the center \( B_{\text{centre}} \) is calculated as:

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

Answer 2

The magnetic field at the center of a square loop is given by the formula: \[ B = \frac{\mu_0 I}{2a} \left( \frac{2}{\pi} \right) \] However, the presence of different resistances for parts ABC and ADC requires us to calculate the net effective current flowing through each section and the resultant magnetic field produced.
For each segment, we calculate their individual contributions based on their respective resistances, and by applying the Biot-Savart law, we arrive at the magnetic field at the center of the loop as: \[ B = \frac{\sqrt{2} \mu_0 I}{3 \pi a} \] Thus, the correct answer is Option 1.