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प्रश्न
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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उत्तर
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
संबंधित प्रश्न
A solenoid of length π m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its center along the axis.
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A solenoid of length π m and 5 cm in diameter has a winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the radius.
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(µ0 = 4π x 10-7 Wb/Am)
