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 (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)*(1.3)/(0.3)`

`\epsilon = 8.6*10^(-3)V`

`\epsilon = 8.6mV`

Therefore the magnitude of average induced emf is 8.6mV