A solenoid of radius 4 cm has 250 turns and a length of 20.0 cm. Find its inductance
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Q: Find the cross-sectional area of a 4.5 cm long solenoid with 515 turns and a self-inductance of 43…
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Q: Find the cross-sectional area of a 5.1 cm long solenoid with 525 turns and a self-inductance of 43…
A: Given data: Length (l) = 5.1 cm = 5.1×10-2 m Number of turns (N) = 525 Self Inductance (L) = 43 μH…
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Q: Consider the RL circuit shown in the figure, where R = 19.8 Q. The switch S is closed at t = 0. At t…
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Q: what isthe current in the circuit after 1 μsec?
A: The equation for the current here is
- When AC current passes through an inductor the magnetic field created will induce an emf inside the inductor which will resist the flow of current. This tendency is called inductance.
- The inductance of the solenoid is directly proportional to the square of the number of turns and cross-sectional area, and it is inversely proportional to the length of the solenoid. It is given by,
Here, μ0 is the permeability of free space, N is the number of turns of the solenoid, A is the area, and l is the length of the solenoid.
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