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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 V_R will be zero at t=0.5 sec

So the option (4) is correct.

Posted by

Ritika Jonwal

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