Question

When a battery is connected to the two ends of a diagonal of a square conductor frame of side $a$, the magnitude of magnetic field at the centre will be ($\mu_0 = $ permeability of free space)

• A. $\frac{\mu_0}{2\pi a}$
• B. $\frac{2\mu_0}{\pi a}$
• C. $\frac{\mu_0}{\pi a}$
• D. zero

Hint:

This is a general symmetry rule in magnetostatics: for any regular closed polygon conductor frame (like an equilateral triangle, square, or ring) where current enters at one node and exits at another, the net magnetic field at the exact geometric center is always zero due to symmetrical branch cancellation! Memorizing this rule saves you from doing long calculations.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A battery introduces current into a square conducting frame at one corner and extracts it from the diagonally opposite corner. We need to compute the net magnetic field vector at the geometric center of this square configuration.

Step 2: Key Formula or Approach:
1. Apply the Biot-Savart Law and the right-hand grip rule to determine the direction of the magnetic field lines produced by each straight wire segment. 2. Use symmetry: when current enters a corner and splits symmetrically across two identical paths, the currents flowing through the branches are exactly equal.

Step 3: Detailed Explanation:
Let the square frame vertices be labeled sequentially as $A, B, C, D$. Suppose the current $I$ enters at vertex $A$ and exits at the diagonally opposite vertex $C$. The current branches into two parallel paths: Path 1 goes through segments $AB$ and $BC$. Path 2 goes through segments $AD$ and $DC$. Since all four sides of the square have equal length $a$ and identical cross-sectional resistance, the total path resistances are perfectly matched ($R_{ABC} = R_{ADC}$). Consequently, the current splits exactly in half: $$I_1 = I_2 = \frac{I}{2}$$ Now look at the direction of the magnetic fields produced at the center point: Currents in path $ABC$ create a magnetic field pointing perpendicularly into the page (by the right-hand rule). Currents in path $ADC$ create a magnetic field pointing perpendicularly out of the page. Because the path geometries, distances to the center, and currents are completely identical, the magnitudes of these opposing fields are equal. Thus, they cancel each other out completely: $$B_{\text{net}} = B_{ABC} - B_{ADC} = 0$$

Step 4: Final Answer:
The net magnetic field at the center is zero, corresponding to option (D).