**(1)** A long straight conductor carries a current that increases at a constant rate. The current induced in the circular conducting loop placed near the straight conductor (fig.) is

(a) zero

(b) constant and in the clockwise direction

(c) constant and in the anticlockwise direction

(d) increasing and in the clockwise direction

(e)** **increasing and in the anticlockwise direction** **

A current flows in the circular loop because of the *increasing *magnetic flux linked with it. This increase in the magnetic flux is to be opposed in accordance with Lenz’s law. The magnetic field lines produced by the straight conductor are directed normally *into* the plane of the circular coil (*away from the reader*). Therefore, the direction of the magnetic field lines produced by the induced current in the circular coil must be directed normal to the plane of the circular coil and *towards the reader*. The induced current should therefore flow in the *anticlockwise* sense.

The magnitude of the induced current is constant since the *rate* of increase of flux is constant. The correct option is (c).

**(2)**In the circuit shown a coil of inductance 0.5 H and negligible resistance is connected to a 12 V battery of negligible internal resistance using a resistive network. If

*I*

_{1}is the current delivered by the battery immediately after closing the switch and

*I*

_{2}is the current (delivered by the battery) long time after closing the switch,

(a) *I*_{1} = 0.5 A and *I*_{2} = 0.6 A

(b) *I*_{1} = 0.6 A and *I*_{2} = 0.6 A

(c) *I*_{1} = 0.5 A and *I*_{2} = 0.5 A

(d) *I*_{1} = 0 A and *I*_{2} = 0.5 A

(e)** ***I*_{1} = 0.6 A and *I*_{2} = 0.5 A** **

The current through the inductance cannot rise abruptly on closing the switch (because of the time constant L/R of the LR circuit). At the moment of switching the circuit on, the circuit behaves as though the inductance branch is absent. The initial current is therefore given by

*I*_{1} = 12 V/ (16+8) Ω = 0.5 A

The current through the inductance rises exponentially with time *t* from zero to the final steady value* I* [in accordance with the equation *I = I*_{0}(1– e^{–}* ^{Rt/L}*) with usual notations]. After a long time, the current through the inductance becomes steady and there is no voltage drop across the inductance. The inductance is in effect short circuited and the final steady current is given by

* I*_{2} = 12 v/(16+4) Ω = 0.6 A

[Note that the parallel combined value of 8 Ω and 8 Ω is 4 Ω]

The correct option is (a).

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