Verified Solution • 2025–26 Session moderate

QPages 249-253: Q4: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q4 Solution: A transmission wire carries a current of 100 A. What would be…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often reach for the finite‑wire expression $B=\frac{\mu_0 I}{4\pi r}(\sin\theta_1+\sin\theta_2)$ and insert $\theta_1=\theta_2=90^\circ$, which yields an extra factor of ½ and a wrong answer. Remember that a transmission line is effectively infinitely long compared to the 8 m distance, so the correct magnitude is $B=\frac{\mu_0 I}{2\pi r}$.

A transmission wire carries a current of 100 A. What would be the magnetic field BB at a point on the road if the wire is 8 m above the road?

Step-by-Step Verified Solution

**Answer:** B=2.5×10−6B = 2.5 \times 10^{-6} T. **Step-by-Step Explanation:** i=100i = 100 A, d=8d = 8 m. B=μ0i2πd=4π×10−7×1002π×8=2.5×10−6B = \frac{\mu_0 i}{2\pi d} = \frac{4\pi \times 10^{-7} \times 100}{2\pi \times 8} = 2.5 \times 10^{-6} T.

All Questions in this Chapter

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