Verified Solution • 2025–26 Session moderate

QPages 249-253: Q23: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q23 Solution: Find the magnetic field at the centre of a rectangular loop of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the half‑length ($l/2$) or half‑width ($b/2$) as the distance $R$ in $B=\frac{\mu_0 I}{4\pi R}(\sin\theta_1+\sin\theta_2)$, but $R$ must be the perpendicular distance from the centre to the wire (i.e., $b/2$ for the long sides and $l/2$ for the short sides). Using the wrong $R$ leads to an incorrect factor and the final $\frac{2\mu_0 I}{\pi}\frac{\sqrt{l^2+b^2}}{lb}$ is missed. Always draw the right‑angle triangle from the centre to the wire ends to obtain both $R$ and the angles correctly.

Find the magnetic field

B

at the centre of a rectangular loop of length

l

and width

b

, carrying a current

i

.

Step-by-Step Verified Solution

**Answer:**

B = \frac{2\mu_0 i}{\pi}\frac{\sqrt{l^2 + b^2}}{lb}

. **Step-by-Step Explanation:** Rectangular loop of length

l

and width

b

. Field at centre due to each pair of opposite sides:

B = \frac{\mu_0 i}{4\pi (b/2)}(\sin\theta_1 + \sin\theta_2) \times 2 + \frac{\mu_0 i}{4\pi (l/2)}(\sin\phi_1 + \sin\phi_2) \times 2

. Simplifying gives the result.

All Questions in this Chapter

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