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Question
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.
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Solution
Given:
Magnitude of current, i = 5 A
Magnetic field intensity, B = 3.14 × 10−2 T
We know that the magnetic field inside a long solenoid having n turns per unit length is given by
`B = mu_0 ni`
`3.14 xx 10^-2 = 4 pi xx 10^-7 xx n xx5`
`⇒ n = (10^-2)/(20xx10^-7)`
`= 5 xx 10^3 = 5000 ` turns / m
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