A solenoid 4cm in diameter and 20cm in length has 250 turns and carries a current of 15A. Calculate the flux through the surface of a disc of 10cm radius that is positioned perpendicular to and centred on the axis of solenoid .
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r = radius of solenoid = 2 cm
L = 20 cm, turns = N = 250
turns per unit length = n = 1,250 turns/meter
current i = 15 A
Magnetic field through the solenoid = μ₀ n i = 4π * 10⁻⁷ * 1,250 * 15 units
B = 2.356 * 10⁻² H
The magnetic field outside the turns of the solenoid is zero, ie., it exists only in the area of π r².
Flux = B * πr² = 0.0185 Volt-sec
L = 20 cm, turns = N = 250
turns per unit length = n = 1,250 turns/meter
current i = 15 A
Magnetic field through the solenoid = μ₀ n i = 4π * 10⁻⁷ * 1,250 * 15 units
B = 2.356 * 10⁻² H
The magnetic field outside the turns of the solenoid is zero, ie., it exists only in the area of π r².
Flux = B * πr² = 0.0185 Volt-sec
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