Verified Solution • 2025–26 Session moderate

QPages 249-253: Q17: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q17 Solution: Figure shows a square loop ABCD with edge-length . The…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the two parallel paths (ABC and ADC) as carrying currents in the same rotational direction, leading to field addition. You must recognize that current entering a parallel junction splits into branches that traverse the loop in opposite senses (clockwise vs counter-clockwise), so the magnetic fields at the center oppose each other and must be subtracted.

Figure shows a square loop ABCD with edge-length

a

. The resistance of the wire ABC is

r

and that of ADC is

2r

. Find the magnetic field

B

at the centre of the loop assuming uniform wires.

Step-by-Step Verified Solution

**Answer:**

B = \frac{3\mu_0 i}{8\sqrt{2} \pi a}

. **Step-by-Step Explanation:** Square loop ABCD with resistance of ABC =

r

and ADC =

2r

. Current divides:

i_{ABC} = \frac{2i}{3}

,

i_{ADC} = \frac{i}{3}

. Field at centre due to each side is

\frac{\mu_0 i}{4\pi (a/2)}(\sin 45� + \sin 45�) = \frac{\mu_0 i}{2\pi a}\sqrt{2}

. Net field:

B = \frac{\mu_0}{2\pi a}\sqrt{2} \left(\frac{2i}{3} - \frac{i}{3}\right) = \frac{\mu_0 i}{6\sqrt{2}\pi a}

.

All Questions in this Chapter

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