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a long solenoid of radius a, carrying n turns per length is looped by a wire with resistance r. a) if the current in the solenoid is increased at constant rate, k, what current flows in the loop? b) which way does it flow? c) if the current i in the solenoid is constant but the solenoid is pulled out of the loop, turned around and reinserted what is the total charge that passes through the resistor?

User Nbrustein
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a. The magnitude of the induced current is proportional to the rate of change of current in the solenoid, the number of turns per unit length and the area and inversely proportional to the resistance of the loop.

b. The induced current in the loop will flow in a counterclockwise direction, if we apply the right hand rule.

c. The total charge that passes through the resistor during this process is Q = ∫I dt

a. When the current in the solenoid increases at a constant rate, it induces an electromotive force (EMF) in the looped wire. The EMF, denoted by ε, is given by:

ε = -μ₀nAkdI/dt

where:

μ₀ is the permeability of free space (4π × 10⁻⁷ H/m)

n is the number of turns per unit length of the solenoid

A is the area enclosed by the loop

k is the rate of change of current in the solenoid (dI/dt)

The induced EMF then causes a current to flow in the loop, according to Faraday's law of induction:

I = ε/R

where:

I is the induced current in the loop

R is the resistance of the loop

we then substitute the expression for ε:

I = (-μ₀nAkdI/dt) / R

b.

The direction of the induced current can be determined using Lenz's law, which states that the induced current always flows in a direction that opposes the change that produced it.

c.

The total charge that passes through the resistor during this process can be calculated using the formula:

Q = ∫I dt

where:

Q is the total charge

I is the induced current

t is the time interval over which the current flows

User Andor Kesselman
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