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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
RELATED QUESTIONS
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.
A toroid of a central radius of 10 cm has windings of 1000 turns. For a magnetic field of 5 × 10-2 T along its central axis, what current is required to be passed through its windings?
What is Solenoid?
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(magnetic field B = 0.2 T, µ0 = 4 x 10-7 SI units)
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A long solenoid has 200 turns per cm and carries a current of 2.5 A. The magnetic field at the center is ______. (µ0 = 4π × 10-7 Wb/m-A)
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For a solenoid and a toroid, the number of turns per unit length is n and the respective interior volume is V. The self inductance is proportional to n2 and V for ______.
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(µ0 = 4π x 10-7 Wb/Am)
