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  • A long solenoid of radius R carries a time(t) - dependent current . A ring of radius 2R is placed coaxially near its middle. During the time inter

A long solenoid of radius R carries a time(t) - dependent current I(t)=I_ot(1-t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0\leq t\leq 1, the induced current (IR) and the induced EMF (VR) in the ring changes as:-
Option: 1 Direction of I_{R} remains unchanged and V_{R} is zero at t = 0.25
Option: 2 Direction of I_{R} remains unchanged and V_{R} is zero at t = 0.5
Option: 3 At t = 0.25 direction of I_{R}  reverses and V_{R}  is maximum  
Option: 4 At t = 0.5 direction of I_{R}  reverses and V_{R}  is zero
 

Answers (1)

best_answer

 

Faraday's law of induction -

 Faraday’s First Law-

Whenever the number of magnetic lines of force (Magnetic Flux) passing through a circuit changes an emf called induced emf is produced in the circuit. The induced emf persists only as long as there is a change of flux.

Faraday’s Second Law-

The induced emf is given by the rate of change of magnetic flux linked with the circuit.

i.e  Rate of change of magnetic Flux= \varepsilon = \frac{-d\phi }{dt}

where d\phi\rightarrow \phi _{2}-\phi _{1}= change in flux

So,

B=\mu _0nI_0t(1-t)

using B_0=\mu _0nI_0

SO B=B_0t(1-t)

Now considering solenoid as ideal solenoid extended up to infinite and ring as its centre

 \phi =BA_s=B_0t(1-t) A_s

e=-\frac{d \phi}{dt} =-B_0A_s (1-2t)

Since induced emf is changing so current will also be changing 

because i=\frac{e}{R}

So, since direction of emf is changing so direction of current is also changing.

And VR will be zero at t=0.5 sec

 

So the option (4) is correct.

Posted by

Ritika Jonwal

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