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As shown in the figure, P and Q are two ...

As shown in the figure, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current `I_(P)` (as seen by E) and an induced current `I_(Q1)` flows in Q. The switch remains closed for a long time. when S is opened, a current `I_(Q2)` flows in Q. Then the direction `IQ_(1) and IQ_(2)` (as seen by E) are

A

respectively clockwise and anti-clockwise

B

both clockwise

C

both anti-clockwise

D

respectively anti-clockwise and clockwise

Text Solution

Verified by Experts

The correct Answer is:
D
When switch `S` is closed magnetic field passing through `Q` increases in the direction from right to left.So, according to Lenz's law induced current in `Q` i.e. `I_(Q)` will flow in such a direction so that the magnetic field lines due to `I_(Q2)` passes from left to right through `Q`.This is possible when `I_(Q1)` flows in anticlockwise direction as seen by `E`.Opposite is the case when switch `S` is opened i.e. `I_(Q2)` will be clockwise as seen by `E`.
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Knowledge Check

  • In fig. A2.27, the charge that flows from P to Q when the switch S is closed is

    A
    `3muC`
    B
    `6muC`
    C
    `9muC`
    D
    `15muC`

  • In the circuit shown in the figure The steady state currents i_(1)and i_(2) in the coils after the switch S is closed are

    A
    `i_(1)=(EL_(2))/(R(L_(1)+L_(2))`
    B
    `i_(1)=(EL_(1))/(R(L_(1)+L_(2))`
    C
    `i_(2)=(EL_(2))/(R(L_(1)+L_(2)))`
    D
    `i_(2)=(Esqrt(L_(1)L_(2)))/(RL_(2))`

  • In the circuit shown in figure switch S is closed at time t=0 Current I from the battery at time t is given by

    A
    `3(1-e^(-2t))`
    B
    `3+e^(-2t))`
    C
    `3(1-e^(-t/9))
    D
    `3-e^(-2t)`