Q. 543.5( 2 Votes )

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.

Answer :

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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