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The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.50 T/s. What is the electric field strength inside the solenoid at a point on the axis? Express your answer as an integer and include the appropriate units.

a) Calculate the electric field strength
b) Determine the direction of the electric field
c) Evaluate the magnetic flux
d) Assess the energy stored in the solenoid

1 Answer

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Final answer:

The electric field strength inside the solenoid at a point on the axis is zero.

Step-by-step explanation:

To calculate the electric field strength inside the solenoid at a point on the axis, we can use Faraday's law of electromagnetic induction, which states that the electric field induced in a closed loop is equal to the rate of change of magnetic flux through the loop. The formula to calculate the induced electric field is given as:

E = (-dφ / dt) / A

Where E is the electric field strength, dφ/dt is the rate of change of magnetic flux, and A is the area of the loop. In this case, the loop is a circular cross-section of the solenoid with a diameter of 5.0 cm.

First, we need to calculate the rate of change of magnetic flux. The formula for magnetic flux through a loop is given as:

Φ = B * A

Where Φ is the magnetic flux, B is the magnetic field strength, and A is the area of the loop. Substituting the given values, we can find the initial magnetic flux through the loop.

To calculate the rate of change of magnetic flux, we need to differentiate the formula with respect to time:

dΦ / dt = B * dA / dt

Since the solenoid is decreasing at a constant rate, the rate of change of the area is zero. Therefore, the rate of change of magnetic flux is also zero. So, the electric field strength inside the solenoid at a point on the axis is zero.

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