Question

A physics student notices that the current in a coil of conducting wire goes from i1 = 0.200 A to i2 = 1.50 A in a time interval of Δt = 0.300 s. Assuming the coil's inductance is L = 2.00 mH, what is the magnitude of the average induced emf (in mV) in the coil for this time interval?

Answer 1

To solve the problem it is necessary to apply the concepts related to Voltage in an Inductor. By general definition the voltage in in an inductor is defined as

$\epsilon = L \frac{di}{dt}$

Where,

$\epsilon =$Voltage induced (emf)

L = Inductance

di = Rate of change of current flow

dt = rate of change of time

Our values are given by

$L = 2mH = 2*10^{-3} H$

$\Delta i = i_2-i_1 = 1.5A-0.2A = 1.3A$

$\Delta t = 0.3s$

Replacing

$\epsilon = 2*10^{-3}*\frac{1.3}{0.3}$

$\epsilon = 8.6*10^{-3}V$

$\epsilon = 8.6mV$

Therefore the magnitude of average induced emf is 8.6mV