CBSE Board Exam 2020: Important MCQs (with Answers) for Class 12 Physics - Chapter 6 – Electromagnetic Induction; Also useful for JEE Main, UPSEE, WBJEE & Other Engineering Entrance Exams
Check important MCQs (with Answers) for Class 12 Physics Board Exam 2020 (Chapter 6 - Electromagnetic Induction). These MCQs are very useful for other competitive exams like UPSEE 2020, JEE Main, WBJEE etc.
CBSE Board Exams 2020 for Class 12 are going to start soon. Many students preparing for CBSE Class 12 Physics Board Exam often ask about important MCQs (Multiple Choice Questions) and in this article, we are going to provide important questions, based on Chapter 6 (Magnetism and Matter) of Class 12 NCERT textbook. Here you will also get important links to access some important articles for the preparation of CBSE 12th board exams 2020.
CBSE Board Exam 2020: Important MCQs (with Answers) for Class 12 Physics, Chapter 6 - Electromagnetic Induction
Q1. The flux linked with a coil at any instant t is given by: ∅ = 2t2 – t + 1. The magnitude of induced emf at t = 3 sec. is:
(a) 3.5 V
(b) 5 V
(c) 8 V
(d) 11 V
Q2. A bar-magnet falls down through a conducting coil as shown in the figure given below. The acceleration of the bar magnet is:
(a) less than g
(b) greater than g
(c) equal to g
Acceleration of the bar magnet will be less than ‘g’.
When the magnet falls down, an induce current is produced in the coil, due to which magnetic field will be produced in such a way that it opposes the freely falling magnet (Lenz’s law).
As magnet falls down the ring, then current will induced in opposite direction, due to which magnetic field will be produced in such a way that it again opposes the free fall of magnet (Lenz’s law).
Q3. The magnetic flux threading a coil changes from 12 Wb to 6 Wb in 1 second. What is the value of induced emf?
(a) 3 volt
(b) ‒ 3 volt
(c) 6 volt
(d) ‒ 6 volt
Given, change in flux = Δ ϕ = 6 Wb ‒ 12 Wb = ‒ 6 Wb, time interval = Δt = 1 second.
Q4. In the figure shown below, M and N are two conducting coils. According to Faraday’s law, the induced emf is given by e = -Δϕ/Δt = - (-6/1) = 6V. What is the shall be the direction of the induced momentary current in M immediately after the switch is closed?
(c) No current will produce
(d) Any random direction
When switch is closed, current in M grows clockwise. The induced current in N will oppose this growth (Lenz’s law). Therefore, current must be in anticlockwise as the loop will be observed from above.
Q5. An stream of electrons is moving along the line AB lying in the same plane as a circular conducting loop. Select the correct options from the following.
(a) A current will be induced in clockwise direction as seen from above
(b) A current will be induced in anticlockwise direction as seen from above
(c) No current will be induced in the coil
(d) A current will be induced and will change direction as electrons move towards and away from the coil
Stream of flowing electrons can be considered as current whose direction is opposite to the flow of direction of electrons.
As electrons are flowing from A to B, it means current is flowing from B to A. Due to this current there will be change in magnetic flux through the coil, so a current will be induced in the coil which oppose the change in flux.
If we observe the coil from above then the direction of current induced in the coil will be in anticlockwise direction.
As the magnetic field in uniform everywhere, so change in magnetic flux linked with the coil is zero.
Therefore, emf induced in the coil is zero.
Q7. Two circulars coil, one of small radius r1and other of very large radius r2 are placed co-axially with centres coinciding. The mutual inductance of the arrangement is:
(d) None of these
If a time varying current I2 is made to flow through the outer circular coil.
The inner coil placed co-axially has very small radius, therefore B2 is considered as constant over its cross sectional are. So, flux associated with inner coil is:
Q8. Three inductors are connected as shown in the figure given below. What is its equivalent inductance?
(a) 0.5 H
(b) 1 H
(c) 1.5 H
(d) 2 H
When two inductors L1and L2 are connected in parallel then their equivalent inductance Lp is given by: 1/LP = 1/L1 + 1/ L2.
When two inductors L1and L2 are connected in series then their equivalent inductance Ls is given by: LS = L1 + L2.
In the given circuit, the inductors of inductance 0.5 H are connected in parallel therefore, 1/LP = 1/0.5 + 1/0.5 = 2/1 + 2/1 = 4H.
This combination is connected with 0.75 H inductor in series therefore, Ls = 0.75 + 0.25 = 1.0 H.
Q9. A copper rod of length l rotates about its end with angular velocity ω in a uniform magnetic filed B. What is the magnitude of emf developed between the ends of the rod?
(d) None of these
AS the rod of length l rotates about its one end then, the area it intercepts is, A = π l2.
The flux linked with this area, d∅ = B πl2
Time taken to complete one rotation = dt = 2π/ω
Q10. A circular conducting coil with radius r and resistance R is placed with its plane perpendicular to a magnetic field varying with time as B = Bo sin ωt. What is the expression for induced current in the loop?