Question

A current of 4 A is passed through a square loop of side 5 cm made of a uniform manganin wire as shown in the figure. The magnetic field at the centre of the loop is

• A. $\frac{24\sqrt{2}}{5} \times 10^{-5}$ T
• B. $\frac{3\sqrt{2}}{5} \times 10^{-5}$ T
• C. $\frac{6\sqrt{2}}{5} \times 10^{-5}$ T
• D. Zero

Hint:

For any closed loop of regular geometric shape made of uniform wire, if current enters at one point and leaves at another point on the loop, the net magnetic field at the center is always zero.

Answer 1

The Correct Option is D

Step 1: Analyze the Circuit:
The figure shows a square loop of uniform wire. Current enters at one corner and leaves at another. Let the side of the square be $a$. The total current is $I$. Let the input terminal be A and the output terminal be B. The current splits into two paths: 1. Short path (one side, length $a$, Resistance $R$). 2. Long path (three sides, length $3a$, Resistance $3R$).
Step 2: Current Division:
Since the wire is uniform, resistance is proportional to length. Current in short path ($I_1$) and long path ($I_2$) divides inversely to resistance. $I_1 = I \left(\frac{3R}{R+3R}\right) = \frac{3}{4}I$. $I_2 = I \left(\frac{R}{R+3R}\right) = \frac{1}{4}I$.
Step 3: Magnetic Field Calculation:
The magnetic field at the center of a square side of length $a$ carrying current $i$ is $B \propto \frac{i}{a}$. Field due to short path (1 side): $B_1 \propto \frac{I_1}{a} = \frac{3I}{4a}$. (Direction: Out of page, say). Field due to long path (3 sides): $B_2 \propto 3 \times \frac{I_2}{a} = \frac{3(I/4)}{a} = \frac{3I}{4a}$. (Direction: Into page).
Step 4: Net Field:
The magnitudes are equal: $B_1 = B_2$. The directions produced by the currents in the two branches at the center are opposite (one clockwise, one counter-clockwise relative to center). Thus, the net magnetic field is $B_{net} = B_1 - B_2 = 0$. Note:
This result holds true for any regular polygon loop of uniform wire where current enters and leaves at any two vertices.