Question

A solenoid with 20 turns/cm has a radius of 2.4 cm. The magnitude of the induced electric field 2cm from the axis is 5x10–3 v/m. At what rate is the current in the solenoid changing

Answers

GPT 3.5
We can use Faraday's law of electromagnetic induction to solve this problem. Faraday's law states that the induced electromotive force (emf) in a closed loop is equal to the negative rate of change of magnetic flux through the loop.

The induced electric field is related to the rate of change of magnetic flux by the equation:
E = -dφ/dt

where E is the electric field, φ is the magnetic flux, and t is time.

The magnetic flux through a solenoid is given by:
φ = BA

where B is the magnetic field and A is the area of the loop.

The area of the loop can be calculated using the formula for the area of a circle:
A = πr^2

The magnetic field inside a solenoid can be calculated using the formula:
B = μ₀nI

where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.

Combining these equations, we can solve for dI/dt:
E = -dφ/dt
5x10^-3 V/m = -d(BA)/dt
5x10^-3 V/m = -d(μ₀nI * πr^2)/dt
5x10^-3 V/m = -μ₀n * dI/dt * πr^2

Rearranging the equation, we get:
dI/dt = -(5x10^-3 V/m) / (μ₀n * πr^2)

We can substitute the given values into this equation to find the rate at which the current is changing.

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