Verified Solution • 2025–26 Session moderate

QPages 249-253: Q18: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q18 Solution: Figure shows a square loop of edge a made of a uniform wire. A…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume the full current $i$ flows through each segment or fail to recognize that the square loop splits into two parallel resistive paths (A-B-C and A-D-C) of equal resistance, causing the current to divide equally into $i/2$ for each path before applying the Biot-Savart law.

Figure shows a square loop of edge a made of a uniform wire. A current ii enters the loop at the point AA and leaves it at the point CC. Find the magnetic field at the point PP which is on the perpendicular bisector of ABAB at a distance a/4a/4 from it.

Step-by-Step Verified Solution

**Answer:** B=μ0i4πa2B = \frac{\mu_0 i}{4\pi a}\sqrt{2}. **Step-by-Step Explanation:** Current enters at A and leaves at C. The loop is divided into two paths: A-B-C and A-D-C. Point P is on the perpendicular bisector of AB at distance a/4a/4. The field is calculated using the Biot-Savart law for each segment.

All Questions in this Chapter

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