Verified Solution • 2025–26 Session moderate

QPages 249-253: Q54: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q54 Solution: The magnetic field inside a long solenoid, carrying a current…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often confuse the total number of turns $N$ with the turn density $n$ (turns per unit length), leading to incorrect units or missing the division by length. Always verify that the final answer is in turns/meter, not just a dimensionless count.

The magnetic field BB inside a long solenoid, carrying a current of 5.00 A, is 3.14×10−2 T3.14 \times 10^{-2} \text{ T}. Find the number of turns per unit length of the solenoid.

Step-by-Step Verified Solution

**Answer:** n=5000n = 5000 turns/m. **Step-by-Step Explanation:** i=5i = 5 A, B=3.14×10−2B = 3.14 \times 10^{-2} T. B=μ0ni⇒n=Bμ0i=3.14×10−24π×10−7×5=5000B = \mu_0 n i \Rightarrow n = \frac{B}{\mu_0 i} = \frac{3.14 \times 10^{-2}}{4\pi \times 10^{-7} \times 5} = 5000 turns/m.

All Questions in this Chapter

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