The magnetic intensity at the center of a long current carrying solenoid is found to be 1.6 × 10³ Am⁻¹. If the number of turns is 8 per cm, then the current flowing through the solenoid is __________ A.

Solution

Physics sir · 3 steps

Magnetic intensity H inside a solenoid is just nI, so μ₀ never enters the calculation.

1. H, not B. Inside a long solenoid the field is B = μ₀ n I. The magnetic intensity is H = B/μ₀, so it reduces to H = nI, measured in A m⁻¹.

   H = nI ⇒ I = H/n

2. Convert turns. n must be in turns per metre. 8 turns per cm is 8 × 100 = 800 m⁻¹.
3. So. I = (1.6 × 10³)/800 = 2 A, so the current is 2 A.

The common slip: 200. Using n = 8 without converting from per cm to per metre gives 200 A, not 2 A.

The idea to keep. Check the unit given: A m⁻¹ means magnetic intensity H = nI, while tesla means B = μ₀ nI. Always put n in turns per metre.

Answer from the NTA final answer key. Solution by PrepWiser.