Asolenoid 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?

induced EMF = 240 V

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

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    ...................1

so E = - L

put here value we get

E = - 8 ×

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

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i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!