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QPages 286-287: Q1: HC Verma: Concepts of Physics - Volume 2 Magnetic Properties of Matter Pages 286-287: Q1 Solution: The magnetic intensity H at the centre of a long solenoid…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often compute $n=H/I$ and leave the result in turns per metre, then forget to convert to turns per centimetre by dividing by 100. Remember $1, ext{m}=100, ext{cm}$, so $750, ext{turns/m}=7.5, ext{turns/cm}$.

The magnetic intensity H at the centre of a long solenoid carrying a current of 2.0 A, is found to be 1500 A m−11500 \text{ A m}^{-1}. Find the number of turns per centimetre of the solenoid.

Step-by-Step Verified Solution

**Answer:** n=7.5 turns/cmn = 7.5 \text{ turns/cm}. **Step-by-Step Explanation:** Given: i=2.0 Ai = 2.0 \text{ A}, H=1500 A m−1H = 1500 \text{ A m}^{-1}. For a long solenoid, H=niH = ni, where nn is the number of turns per unit length.
n=Hi=15002=750 turns/m=7.5 turns/cmn = \frac{H}{i} = \frac{1500}{2} = 750 \text{ turns/m} = 7.5 \text{ turns/cm}

All Questions in this Chapter

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