Question

A physics student notices that the current in a coil of conducting wire goes from in 0.200 A to 12 = 1.50 A in a time interval of At = 0.250 s. Assuming the coil's inductance is L = 3.00 mt, what is the magnitude of the average induced emf (in mV) in the coil for this time interval?

Answer 1

Step-by-step explanation:

We can use Faraday's law of electromagnetic induction to find the average induced emf in the coil. According to this law, the induced emf (ε) in a coil is equal to the negative of the rate of change of magnetic flux through the coil:

ε = - dΦ/dt

where Φ is the magnetic flux through the coil.

The magnetic flux through a coil of inductance L is given by:

Φ = LI

where I is the current in the coil.

Differentiating both sides of this equation with respect to time, we get:

dΦ/dt = L(dI/dt)

Substituting the given values, we get:

dI/dt = (1.50 A - 0.200 A) / 0.250 s = 4.40 A/s

L = 3.00 mH = 0.00300 H

Therefore, the induced emf in the coil is:

ε = - L(dI/dt) = - (0.00300 H)(4.40 A/s) = -0.0132 V

Since the question asks for the magnitude of the induced emf, we take the absolute value of the answer and convert it from volts to millivolts:

|ε| = 0.0132 V = 13.2 mV

Therefore, the magnitude of the average induced emf in the coil for the given time interval is 13.2 mV.