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Question
A solenoid of length 50 cm of the inner radius of 1 cm and is made up of 500 turns of copper wire for a current of 5 A in it. What will be the magnitude of the magnetic field inside the solenoid?
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Solution
The magnitude of the magnetic field inside the solenoid,
B = `mu_0"N"/"l""i" = 4pi xx 10^-7 xx 500/0.5 xx 5 = 6.284 xx 10^-3`T
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