A long, ideal solenoid has a diameter d=12cm and n=1200turns/meter carrying current I=20A. If the current is lowered at 4.33amp/s to zero, what is the magnitude of the induced electric field in V/m at a position 8.2cm from the solenoid's axis

To find the magnitude of the induced electric field at a position 8.2cm from the solenoid's axis, we can use Faraday's law of electromagnetic induction.

Faraday's law states that the magnitude of the induced electric field is equal to the rate of change of magnetic flux through a closed loop divided by the area within the loop. In this case, the loop is a circle with a radius of 8.2cm.

To calculate the rate of change of magnetic flux, we need to know the magnetic field inside the solenoid. For an ideal solenoid, the magnetic field inside can be approximated by:

B = μ₀ * n * I

Where:
B is the magnetic field
μ₀ is the permeability of free space (constant)
n is the number of turns per unit length
I is the current

We know the diameter of the solenoid is 12cm, so the radius is 6cm. We can convert this to meters by dividing by 100. Therefore, the radius is 0.06m.

The area of the circle within the loop is given by A = π * r². Plugging in the value of the radius (0.082m), we get A = π * (0.082)^2.

Now, let's calculate the rate of change of magnetic flux:

dΦ/dt = B * A

To find the induced electric field, we need to divide this by the area within the loop, which is A.

E = (dΦ/dt) / A

Finally, substituting the values we have:

B = μ₀ * n * I
A = π * (0.082)^2
E = (B * A) / A

Now we can calculate the value for E.