HOTs Electromagnetic Induction Class 12 Physics

Please refer to Electromagnetic Induction HOTs Class 12 Physics provided below with Electromagnetic Induction. All HOTs for Class 12 Physics with answers provided below have been designed as per the latest syllabus and examination petter issued by CBSE, NCERT, KVS. Students of Standard 12 Physics should learn the solved HOTS for Class 12 Physics provided below to gain better marks in examinations.

Electromagnetic Induction Class 12 Physics HOTs

Assertion (A) & Reason(R)

A – if both Assertion and Reason — are true and Reason — is correct explanation of the Assertion.
B – if both Assertion and Reason — are true but Reason — is not correct explanation of Assertion.
C – if Assertion is true but Reason — is false.
D – if both Assertion and Reason — are false.
E – if Assertion is false but Reason — is true

Question. Assertion– The mutual induction of two coils is doubled, if the self-inductance of the primary or secondary coil is doubled
Reason — Mutual induction is proportional to self-inductance of primary and secondary coils.

Question. Assertion- Making and breaking of current in a coil produce no momentary current in the neighboring coil of another circuit
Reason — Momentary current in the neighboring coil of another circuit is an eddy current.

Question. Assertion- If primary coil is connected by voltmeter and secondary coil by ac source. If large (2020- copper sheet is placed between two coils, induced emf in primary coil is reduced
Reason — Copper sheet between coils has no effect on induced emf in primary coil.

Question. Assertion- An electric motor will have maximum efficiency when back emf becomes equal to half of applied emf
Reason — Efficiency of electric motor depends only on magnitude of back emf.

Question. Assertion- Armature current in DC motor is maximum when the motor has just started
Reason — Armature current is given by I=E-e/R where e is back emf, R is resistance of armature/

Question. Assertion- Eddy current is produced in any metallic conductor when magnetic flux is changed around it
Reason — Electric potential determine the flow of charge.

Question. Assertion — The quantity L/R possesses dimensions of time
Reason — to reduce the rate of increase of current through a solenoid should increase the time constant L/R.

Question. Assertion- Faraday laws are consequence of conservation of energy
Reason — In a purely resistive AC circuit, the current lags behind the emf in phase.

Question. Assertion- Only a change in magnetic flux through a coil maintain a current in the coil if the current is continues
Reason — The presence of large magnetic flux through a coil maintain a current in the coil if the current is continues.

Question. Assertion – magnetic flux can produce induced emf
Reason — Faraday established induced emf experimentally.

Case Study Based Questions

Question. An inductor is simply a coil or a solenoid that has a fixed inductance. It is referred to as a choke. The usual circuit notation for an inductor is as shown.

Let a current i flows through the inductor from A to B. Whenever electric current changes through it, a back emf is generated. If the resistance of inductor is assumed to be zero (ideal inductor) then induced emf in it is given by
e=V_B-V_A = – L di / dt
Thus, potential drops across an inductor as we movein the direction of current. But potential also drops across a pure resistor when we move in the direction of the current.
The main difference between a resistor and an inductor is that while a resistor opposes the current through it, an inductor opposes the change in current through it.
Now answer the following questions.
(1) How does inductor behave when
(a) a steady current flow through it?
(b) a steadily increasing, current flows through it?
(c) a steadily decreasing current flows through it?
(d) Name the phenomenon in which change in current in a coil induces EMF in coil itself?
Answer. (i) (a) As electric current is steady therefore
di / dt = 0;
:: induced emf = e = 0 and the inductor behaves as short circuit.
(b) in the expression
e= – L di / dt
as di / dt is positive EMF is negative. that is VB < VA. That is back EMF is genreted that opposses the increase in current. (c ) di / dt is negative, therefore EMF is positive. that is VB > VA. Forward EMF is generated that opposses fall in current.
(d) Self induction.

Question.
(a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. Can we hope to generate current in the loop by using very strong magnets?
(b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop
(i) when it is wholly inside the region between the capacitor plates
(ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop.
(c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (Figure) to a field-free region with a constant velocity v. In which loop do you expect the induced emſ to be constant during the passage out of the field region? The field is normal to the loops.

(d) Predict the polarity of the capacitor in the situation described by the figure

Answer:
(a) No. However strong the magnet may be current can be induced only by changing the magnetic flux through the loop.
(b) No current is induced in either case. Current can not be induced by changing the electric flux.
(c) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly,
(d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor.

Question.
Given figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12 cm s^-1 in the direction shown.
Give the polarity and magnitude of the induced emf.

(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed
(=12 cm/ sec) when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Answers:
(a) EMF = vBL = 0.12 0.50 x 0.15 = 9.0 mV;
P positive end and Q negative end.
(b) Yes. When K is closed, the excess charge is maintained by the continuous flow of current.
(c) Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite signs at the ends of the rod.
(d) Retarding force = IBL
9 mV / 9 mΩ x 0.5 T x 0.15 m
= 75 x 10-3 N
e) Power expended by an external agent against the above retarding force to keep the rod moving uniformly at 12 cm s’
= 75 x 10-3 x 12 x 10-2 = 9.0 x 10-3 W
When K is open, no power is expended.
(f) I2 R = 1x1x 9 x 10-3 = 9.0 x 10-3 W
The source of this power is the power provided by the external agent as calculated above. g) Zero: motion of the rod does not cut across the field lines. [Note: length of Pg has been considered above to be equal to the spacing between the rails.]

Question. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step up transformer at the plant.
Answers.
Line resistance = 30 X 0.5 = 15Ω rms current in the line . 800 x 1000 W / 4000 V = 200 A
(a) Line power loss = (200 A)2 x 15 Ω = 600 kW.
(b) Power supply by the plant = 800 kW + 600 kW = 1400 kW.
(c) Voltage drop on the line = 200 A 15Ω = 3000 V. The step-up transformer at the plant is 440 V – 7000V

Question. Electromagnetic induction is defined as the production of an electromotive force across an electric conductor in the changing magnetic field. The discovery of induction was done by Michael Faraday in the year 1831.
Electromagnetic induction finds many applications such as in electrical components which includes transformers, inductors, and other devices such as electric motors and generators.
Alternating current is defined as an electric current which reverses in direction periodically. In most of the electric power circuits, the waveform of alternating current is the sine wave.

Question. Consider an inductor whose linear dimensions are tripled and the total number of turns per
unit length is kept constant, what happens to the self-inductance?
(a) 9 times
(b) 3 times
(c) 27 times
(d) 13 times

Question. How to increase the energy stored in an inductor by four times?
(a) By doubling the current
(b) This is not possible
(c) By doubling the inductance
(d) By making current 2–√ times

Question. What will be the acceleration of the falling bar magnet which passes through the ring such that the ring is held horizontally and the bar magnet is dropped along the axis of the ring?
(a) It depends on the diameter of the ring and the length of the magnet
(b) It is equal due to gravity
(c) It is less than due to gravity
(d) It is more than due to gravity

Question. Lenz law is based on which of the following conservation>
(a) Charge
(b) Mass
(c) Momentum
(d) Energy