Question

The magnetic field of a cylindrical magnet that has a pole-face diameter of 3.3 cm can be varied sinusoidally between 29.6 T and 30.0 T at a frequency of 17 Hz. (The current in a wire wrapped around a permanent magnet is varied to give this variation in the net field.) At a radial distance of 2.5 cm, what is the amplitude of the electric field induced by the variation

Answer 1

The amplitude of the electric field induced by the variation in the magnetic field at a radial distance of 2.5 cm is approximately 106.1 V.

Step-by-step explanation:

To determine the amplitude of the electric field induced by the variation in the magnetic field at a radial distance of 2.5 cm, we can use Faraday's law of electromagnetic induction.

Faraday's law states that the induced emf in a conducting loop is equal to the rate of change of magnetic flux through the loop.

The magnetic field can be modeled as a sinusoidal function, so the rate of change is equal to the frequency times the difference between the maximum and minimum magnetic field values. Therefore, the induced emf is given by:

E = 2πrfBmax, where E is the induced emf, r is the radial distance, f is the frequency, and Bmax is the maximum magnetic field.

Using the given values, we can calculate the induced emf:

E = 2π(2.5 cm)(17 Hz)(30.0 T - 29.6 T) = 2π(2.5 cm)(17 Hz)(0.4 T) ≈ 106.1 V

The amplitude of the induced electric field is equal to the amplitude of the induced emf, so the amplitude of the electric field induced by the variation in the magnetic field at a radial distance of 2.5 cm is approximately 106.1 V.

Answer 2

As we know that induced EMF is given by rate of change in magnetic flux

so here we can say

`EMF = (d\phi)/(dt)`

here we know that

`\phi = BA`

so here we can say

`EMF = A(dB)/(dt)`

`EMF = A*(\Delta B)* f`

now by relation of EMF and electric field we can say

`\int E. dl = EMF`

`E. 2 \pi r = \pi r^2 *(\Delta B)* f`

`E = (r)/(2) * (\Delta B) * f`

now plug in all values

r = 2.5 cm

`\Delta B = 30.0 - 29.6 = 0.4 T`

f = 17 Hz

`E = (0.025)/(2) * 0.4 * 17 = 0.085 V/m`

so electric field is given as E = 0.085 N/C