Question

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QUESTION :-
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A long solenoid of radius 0.01 has 600 turns per metre a small coil of 100 turns is wrapped closely around the centre of the bigger solenoid. If the current in the solenoid Rises from 0 to 2 ampere in 1 ms. Find the induced EMF in the smaller Coil.
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Answer 1

Answer:

The EMF induced in the smaller coil is - 150 V

According to the problem the solenoid has the radius ,r = 0.01

and it has 600 turns /m

Now if the current is 2A passing through the solenoid

Therefore,the magnetic field produced is

B=μ0nI [ μ= coefficient, n=number of turns and l is the length]

    =1.25×10^(−6)×600×2

       =0.0015 T

So the rate of change of magnetic field is 0.0015−0/1×10^(−3)

                                                                  =1.5 T/s

Now induced emf is given by EMF = −NdB/dt =−100×1.5

                                                                            =−150 V

Please mark it brainliest

Answer 2

Answer:Given that,R=0.01,n=600turns

Change in current is di=2-0=2amperes,t=1×10–3s

First find B

B=uoni

B=1.25×10–6×600×2=1.25×10-6×1200

B=0.0015=15×10–4Tesla

Now change in magnetic field density is,

dB/dt=15×10-4-0/10-3=15×10–1

dB/dt=1.5T/s

Hence,EMF=-NdB/dt=-100×1.5=-150v

So electromotive force is -150v

Explanation:

Hope u can understand it frnd........