# The magnetic field B inside a long solenoid, carrying a current of 5.00 A, is 3.14 × 10–2 T. Find the number of turns per unit length of the solenoid.

Given:

Current(i) = 5 A

Magnetic field(B) = 3.14 × 10–2 T

Formula used:

Magnetic field inside a long solenoid B = μ0ni,

Where

μ0 = magnetic permeability of vacuum = 4π x 10-7 T m A-1, n = number of turns per unit length, i = current carried by the wire

Hence, from the given information:

3.14 × 10–2 = 4π x 10-7 x n x 5

=> n = 5000 turns/m

Therefore, number of turns per unit length of the solenoid = 5000(Ans)

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