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 (so at a point outside the solenoid)?
To find the magnitude of the induced electric field at a position outside the solenoid, 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 passing through a surface.
To calculate the magnetic flux, we need to know the magnetic field inside the solenoid, which can be determined using the formula:
B = μ₀ * n * I
Where B is the magnetic field, μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), n is the number of turns per unit length, and I is the current.
The magnetic field inside the solenoid is uniform and directed along its axis.
In this case, the magnetic field, B, can be calculated as follows:
B = μ₀ * n * I
B = (4π × 10⁻⁷ T·m/A) * (1200 turns/m) * (20 A)
Next, we need to calculate the change in magnetic flux, ΔΦ. Since the solenoid's diameter is given, we can use the formula for the area of a circle to find the area of the surface through which the flux passes.
A = π * r²
Where A is the area and r is the radius of the circle.
In this case, A = π * (0.12/2)² = π * 0.06²
Now, the change in magnetic flux can be calculated as:
ΔΦ = B * ΔA
Where ΔA is the change in area.
Finally, the induced electric field is given by:
E = - (dΦ/dt) / ΔA
where dΦ/dt is the rate of change of magnetic flux and ΔA is the area.
In this case, we are given the rate of change of current, which is 4.33 A/s. The area ΔA can be calculated as the difference between the areas of two concentric circles, one with a radius of 8.2 cm and the other with a radius of 6 cm as they represent surfaces through which the flux passes.
Using these formulas, you can substitute the given values and calculate the magnitude of the induced electric field, E.