Question

A solenoid with an inductance of 8 mH is connected in series with a resistance of 5 Ω and an EMF forming a series RL circuit. A current of 4.0 A is flowing clockwise in the solenoid at time t = 0. The current in the solenoid then grows from 4.0 A to 10.0 A in 0.2 ms. What are the value and the direction of the induced EMF?

Answer 1

induced EMF = 240 V

and by the lenz's law direction of induced EMF is opposite to the applied EMF

Step-by-step explanation:

given data

inductance = 8 mH

resistance = 5 Ω

current = 4.0 A

time t = 0

current grow = 4.0 A to 10.0 A

to find out

value and the direction of the induced EMF

solution

we get here induced EMF of induction is express as

E = - L
`(dI)/(dt)` ...................1

so E = - L
`(I2 - I1)/(dt)`

put here value we get

E = - 8 ×
`10^(-3)`
`(10 - 4)/(0.2*10^(-3))`

E = -40 × 6

E = -240

take magnitude

induced EMF = 240 V

and by the lenz's law we get direction of induced EMF is opposite to the applied EMF