# Solution of Hots

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HOTS - 1

1. No induced emf will be induced at the terminals of the coil as there will be no change in magnetic flux linked with it as the magnet rotates about its own axis.
2. An emf will be induced at the terminals of the coil as magnetic flux linked with it changes due to the rotation of the magnet about an axis perpendicular to its length.

HOTS - 2

The variation of magnetic flux and induced emf with time is shown in figure H - 2. There is no change in magnetic flux until the right side of the coil enters the field at time to. Thereafter, the magnetic flux increases linearly till the coil is completely inside the field at time t1.

There is no change in flux from t1 to t2, the time interval in which the coil moves completely inside the field. At time t2, the coil begins to come out of the field. Thereafter, the magnetic flux decreases linearly to zero till the coil is completely out of the field at time t3. The variation (increase and decrease) in magnetic flux is linear for a rectangular coil.

HOTS - 3

When the switch is closed, a current is set up in the rod from Q to P. According to Fleming's left hand rule, a force acts on the rod towards right. This accelerates the rod towards right. As it speeds up, the motional emf produced in the rod starts to oppose the emf of the battery. Finally, the two emfs cancel out each other and hence, the current flowing in the rod becomes zero. Thereafter, the rod continues to move with constant speed.

HOTS - 4

An emf is induced in a conductor only if its length is perpendicular to both velocity and magnetic field. Therefore, emf will be induced in AD, BC, EH and GF and it will be given by vBL.

HOTS - 5

In a dc circuit, the role of an inductor is to delay the growth of current due to the induced emf. Therefore, the current in the bulb B1 gets delayed and it lights up later as compared to the bulb B2. Yes, the two bulbs will be equally bright after some time when the induced emf becomes zero as resistance of L and R is same and inductance has no effect.

HOTS - 6

In a dc circuit, the role of an inductor is to delay the growth of current due to the induced emf. Therefore, the current in the bulb B1 gets delayed and it lights up later as compared to the bulb B2. No, the bulb B1 will be more bright after long time when the induce emf becomes zero. This is due to the fact that the resistance of L is zero (ideal inductor) and the net resistance in the branch of B1 is less (bulb only) than the net resistance in the branch of B2 (bulb and R).

CBSE Electromagnetic Induction ( With Hint / Solution)
Class XII (By Mr. Ashis Kumar Satapathy)
email - [email protected]

Electromagnetic Induction