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Electromagnetic Induction MSQ


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10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Electromagnetic Induction MSQ

Electromagnetic Induction MSQ for IIT JAM 2023 is part of Topic wise Tests for IIT JAM Physics preparation. The Electromagnetic Induction MSQ questions and answers have been prepared according to the IIT JAM exam syllabus.The Electromagnetic Induction MSQ MCQs are made for IIT JAM 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Electromagnetic Induction MSQ below.
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*Multiple options can be correct
Electromagnetic Induction MSQ - Question 1

If a conductor 0.2m long moves with a velocity of 0.3m/s in a magnetic field of 5T, calculate the emf induced if magnetic field, velocity and length of conductor are mutually perpendicular to each other. 

Detailed Solution for Electromagnetic Induction MSQ - Question 1

The formula for induced emf is: emf = Blv if B, l, v are perpendicular to each other. Substituting the values of B, l and v from the question, we get emf = 0.3V.

*Multiple options can be correct
Electromagnetic Induction MSQ - Question 2

A constant force F is being applied on a rod of length ‘l’  kept at rest on two parallel conducting rails connected at ends by resistance R in uniform magnetic field B  as shown. Which of the following statement is incorrect.

Detailed Solution for Electromagnetic Induction MSQ - Question 2


as ‘a’  is decreasing continuously hence the rate of power delivered by external force will be decreasing continuously.

The correct answers are: the power delivered by force will be constant with time, the power delivered by force will be increasing first and then will decreases, the rate of power delivered by the external force will be increasing continuously

*Multiple options can be correct
Electromagnetic Induction MSQ - Question 3

In the figure shown ‘R’  is a fixed conducting fixed ring of negligible resistance and radius ‘a’PQ  is a uniform rod of resistance r. It is hinged at the centre of the ring and rotated about this point in clockwise direction with a uniform angular velocity ω. There is a uniform magnetic field of strength ‘B’  pointing inwards ‘r’  is a stationary resistance

Detailed Solution for Electromagnetic Induction MSQ - Question 3

Equivalent circuit :


By nodal equation :

nodal



also direction of current in ‘r’ will be towards negative terminal i.e. from rim to origin.
Alternatively; by equivalent of cells (figure (ii)):

The correct answers are: Current through ‘r’  is  Direction of current in external ‘r’  is from circumference to centre

*Multiple options can be correct
Electromagnetic Induction MSQ - Question 4

In the figure shown, the magnet is pushed towards the fixed ring along the axis of the ring and it passes through the ring. Which of the following is incorrect.

*Multiple options can be correct
Electromagnetic Induction MSQ - Question 5

In the circuit diagram shown

Detailed Solution for Electromagnetic Induction MSQ - Question 5

At steady state no resistance will be offered by inductor hence, time constant is L/R  and steady state current in inductor is ε/R. 

The correct answers are: time constant is  L/R , steady state current in inductor is  ε/R

*Multiple options can be correct
Electromagnetic Induction MSQ - Question 6

An ideal inductor, (having initial current zero) a resistor and an ideal battery are connected in series at time t = 0. At any time t, the battery supplies energy at the rate PB, the resistor dissipates energy at the rate PR and the inductor stores energy at the rate PL.

Detailed Solution for Electromagnetic Induction MSQ - Question 6

By principal of energy conservation  
Near the starting of the circuit 
As di/dt has greater value at the starting of the circuit, 
The correct answers are:  for all times tonly near the starting of the circuit

*Multiple options can be correct
Electromagnetic Induction MSQ - Question 7

Two straight conducting rails form a right angle where their ends are joined. A conducting bar in contact with the rails starts at the vertex at time t = 0 and moves with constant velocity v along them as shown in figure. A magnetic field   is directed into the page. The induced emf  in the circuit at any time t  is proportional to

Detailed Solution for Electromagnetic Induction MSQ - Question 7




The correct answers are: tv2

*Multiple options can be correct
Electromagnetic Induction MSQ - Question 8

A conducting rod of length l is moved at constant velocity v0 on two parallel, conducting, smooth, fixed rails, that are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown in figure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct:

Detailed Solution for Electromagnetic Induction MSQ - Question 8

Rate of work done by external agent is :

and thermal power dissipated in the resistor = eI = (BvL)I

clearly both are equal.

If applied external force is doubled, the rod will experience a net force and hence acceleration. As a result velocity increases.
Since I = e/R

On doubling ‘R’, current and hence required power becomes half.

Since, P = BILv

The correct answers are: The thermal power dissipated in the resistor is equal to rate of work done by external person pulling the rod, If applied external force is doubled than a part of external power increases the velocity of rod, If resistance R is doubled then power required to maintain the constant velocity v0 becomes half

*Multiple options can be correct
Electromagnetic Induction MSQ - Question 9

A circuit consisting of a constant emf‘E’, a self induction ‘L’ and a resistance ‘R’  is closed at  t = 0. The relation between the current I in the circuit and time t  is as shown by curve ‘a’  in the figure. When one or more of parameters E, R and L are changed, the curve ‘b’ is obtained. The steady state current is same in both the cases. Then it is possible that

Detailed Solution for Electromagnetic Induction MSQ - Question 9

Since curve ‘b’  is below curve ‘a’  hence it is possible that   and  R  are kept constant and L is increased or  E  and R are both halved and  L  is kept constant.

The correct answers are: E and are kept constant and L is increased , E  and R are both halved and L is kept constant

*Multiple options can be correct
Electromagnetic Induction MSQ - Question 10

A conducting loop rotates with constant angular velocity about its fixed diameter in a uniform magnetic field. Whose direction is perpendicular to that fixed diameter.

Detailed Solution for Electromagnetic Induction MSQ - Question 10

The correct answers are: The emf  will be maximum at the moment when flux is zero, The emf  will be zero at the moment when flux is maximum, The emf  will be maximum at the moment when plane of the loop is parallel to the magnetic field, The phase difference between the flux and the emf  is π/2

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