A long solenoid of radius a carrying n turns per unit lengthis looped by a wire of resistance R, as shown in the figure.If the current I(t) in the solenoid is increasing at a constantrate, dI / dt = h = constant,(a) What is the magnitude of the current that flowsthrough the loop?(b) If the current I in the solenoid is constant but thesolenoid is pulled out of the loop and reinserted in the opposite direction, what total charge passesthrough the resistor?(c) While the current is changing in the wire that is wrapped around the solenoid, what is the directionand magnitude of the electric field inside the solenoid?
Question
A long solenoid of radius a carrying n turns per unit lengthis looped by a wire of resistance R, as shown in the figure.If the current I(t) in the solenoid is increasing at a constantrate, dI / dt = h = constant,(a) What is the magnitude of the current that flowsthrough the loop?(b) If the current I in the solenoid is constant but thesolenoid is pulled out of the loop and reinserted in the opposite direction, what total charge passesthrough the resistor?(c) While the current is changing in the wire that is wrapped around the solenoid, what is the directionand magnitude of the electric field inside the solenoid?
Solution
(a) The magnitude of the current that flows through the loop can be found using Faraday's law of electromagnetic induction. This law states that the induced electromotive force (EMF) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.
The magnetic field inside a solenoid is given by B = μ₀nI, where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. The magnetic flux Φ through the loop is then given by Φ = BA = μ₀nIA, where A is the area of the loop (πa² for a loop of radius a).
The rate of change of the flux is then dΦ/dt = μ₀nA(dI/dt) = μ₀nπa²h, where h = dI/dt is the rate of change of the current.
The induced EMF is then given by |ε| = dΦ/dt = μ₀nπa²h.
The current I' that flows through the loop is then given by Ohm's law, I' = ε/R = μ₀nπa²h/R.
(b) If the current I in the solenoid is constant but the solenoid is pulled out of the loop and reinserted in the opposite direction, the magnetic flux through the loop changes by 2Φ (once when the solenoid is removed, and once when it is reinserted). The total charge Q that passes through the resistor is then given by Q = ∫I'dt = 2εt/R = 2μ₀nπa²ht/R.
(c) While the current is changing in the wire that is wrapped around the solenoid, the direction of the electric field inside the solenoid is along the length of the solenoid, and its magnitude E can be found from Faraday's law. The EMF around a loop of length l inside the solenoid is given by ε = El, so E = ε/l = μ₀nπa²h/l.
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