WBCHSE Class 12 Physics Electromagnetic Induction Questions And Answers

Electromagnetic Induction Long Questions And Answers

Question 1. A cylindrical bar magnet Is placed along the axis of a circular coll. If the magnet is rotated about that axis, will any current be induced in the coil?

Answer:

Since the bar magnet is cylindrical, its lines of force remain symmetric about its axis. As a result, if the bar magnet is rotated about its axis, i.e., for the axis of the coil, the number of magnetic lines of force Le., magnetic flux linked with the coil remains unchanged. The induced in the coil.

Question 2. Concerning where a small bar magnet M approaches a coil C (with ends connected to a galvanometer), show that Lenz’s law is consistent with the principle of conservation of energy. What will be the Magna polarity of the coil according to the diagram, as the magnet moves the night and comes out from the left?

Answer:

According to Lenz’s law, the direction of induced current in the coil will be such that, the end Q opposes the motion of the magnet. Hence, a north pole is generated at the end Q, i.e., the direction of current is anticlockwise.

To continue the motion of the magnet against the repulsive force between the magnet and the coil, some external work must be done.

This external work is converted into electrical energy in the coil. Hence, Lenz’s law is the law of conservation of energy.

If the magnet moves through the coil and comes out from the left, the induced current in the coil again opposes that motion, i.e., the end P attracts the S pole.

So, the direction of the current in the coil will be reversed. This event can be noticed with the help of the galvanometer G.

WBBSE Class 12 Electromagnetic Induction Q&A

Question 3. A conducting wire is bent in the form of a circle and a straight conductor AB is kept outside but near the conducting circle. The wires are in the plane of the paper. If the current flowing from A to B gradually increases in magnitude, will there be any current in the circular conductor? If so, in what direction?

Answer:

A magnetic field exists around the current-carrying conductor AB. The magnetic lines of force thus generated around AB) cut the plane of the circular wire.

With the increase in current, the magnetic field as well as the number of lines of force linked with the circular wire increases.

So, current is induced in the circular conductor; and its direction is such that the circular conductor tries to resist the increase of current in the wire AB (according to Lenz’s law).

This can occur only if the induced current flows in a clockwise direction (viewed from the upper face) in the circular conductor.

Question 4. Two conducting rings P and Q of radius r and 3r move in opposite directions with velocities 2v and v respectively on a conducting surface S. There Is a uniform magnetic field of magnitude B perpendicular to the plane of the rings. What Is the potential difference between the highest points of the two rings?

Answer:

The potential difference between highest point E and lowest point A of ring P,

⇒ \(V_A-V_E=B \cdot 2 v \cdot 2 r=4 B v r\) → (1)

A potential difference between highest point D and lowest point C of ring Q,

⇒ \(V_D-V_C=B \cdot v \cdot 6 r=6 B v r\) → (2)

As the two rings move on the same conducting surface S,

∴ \(V_A=V_C\)

From equations (1) and (2),

∴ \(V_D-V_E=10 B v r\)

Question 5. Two identical circular colls A and B are placed parallel to each other with their centres on the same axis. The coil B carries current in the clockwise direction as seen from A. What would be the direction of the induced current in A as seen from B when

1. The current in B is increased,
2. The coil B is moved towards A, keeping the current in B constant.

Answer:

1. If the current in coil B is increased, a current will be induced in coil A and according to Lenz’s law, its direction will be such that, it will oppose the increase of current in coil B. Hence the direction of induced current in coil A will be opposite to the direction of current in coil B. So, when looking from coil B, the current in coil A will be clockwise.
2. Keeping the current steady in coil B, if it is moved towards coil A, the number of magnetic lines’ 6f force linked with coil A will increase and current to that in coil B will be induced in coil A. Hence, in this case, if looked at from coil B, the current in coil A will be clockwise.

Key Concepts in Electromagnetic Induction

Question 6. Show that the units of RC and \(\frac{L}{R}\) arc of time. R, L and C carry their usual significance.

Answer:

Unit of RC = \(\Omega \times F=\frac{V}{A} \times \frac{C}{V}=\frac{C}{A}=\frac{C}{\underline{S}}=s\)

So, the unit of RC and the unit of time are the same.
Again, according to the law of electromagnetic induction,

⇒ \(e=-L \frac{d I}{d t}\)

So, \(V=H \times \frac{A}{s} \quad \text { or, } \frac{V}{A} \times s=H\)

Or, \(\Omega \times \mathrm{s}=\mathrm{H} \quad \text { or, } \frac{\mathrm{H}}{\Omega}=\mathrm{s}\)

So, the unit of \(\frac{L}{R}\) and the unit of time are the same.

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Question 7. Three identical closed coils A, B and C are placed parallel. Coils A and C carry equal currents as shown. Coils B and C are fixed and coil A is moved towards B with uniform speed. Will there be any induced current in B? If no, give a reason. If yes, mark the direction of the induced current in

Answer:

As coil A moves towards coil B, the number of magnetic lines of force linked with coil B will increase and hence a current will be induced in coil B.

Since there is no relative motion between coils B and C, the current in coil C cannot induce any current in coil B. So, according to Lenz’s law, the direction of current in coil B will be opposite to that in A, is depicted

Question 8. A coil of resistance R having n turns is connected with a galvanometer of resistance 4R. If the magnetic flux changes from Φ₁ to Φ₂ in time t in the circuit, what will be the value of the induced current?

Answer:

Effective resistance = R + 4R = 5R

For n turns the effective change in the magnetic flux = \(n\left(\phi_2-\phi_1\right)\)

So, the emf induced = \(-\frac{n\left(\phi_2-\phi_1\right)}{t}\)

∴ Induced current = \(-\frac{n\left(\phi_2-\phi_1\right)}{5 R t}\)

9. Self-Inductance of a coll is 2mH; current through this coll is, I = t² e^-t (t = time). After how much time will the Induced emf be zero?

Answer:

Induced emf \(E=-L \frac{d I}{d t}\)

Now, \(\frac{d I}{d t}=\frac{d}{d t}\left(t^2 e^{-t}\right)=2 t e^{-t}-t^2 e^{-t}\)

⇒ \(-e^{-t} \cdot t(t-2)\)

Hence, \(E=L e^{-t} t(t-2)\)

So, if E = 0, then t = 0 or 2

Therefore, after 2 s from the initial moment, the induced emf will be zero.

Question 10. The north pole of a bar magnet faces a closed circular coil. It oscillates rapidly along the common axis of the magnet and the coil. Determine the direction of induced current in the coil from Lenz’s law.

Answer:

When the north pole of the bar magnet comes nearer the coil, then according to Lenz’s law, the coil repels that pole. So, at the end of the coil facing the magnet, a north pole will be generated due to induced current.

When viewed from the side of the magnet, this current will appear anticlockwise in direction. Conversely, when the magnet is taken away from the coil, the induced current at the same end will be clockwise in direction.

If the magnet is oscillated periodically the induced current in the coil will also change direction periodically. Thus, an alternating current will be induced in the coil.

Short Answer Questions on Faraday’s Law

Question 11. A bar magnet is pulled through a conducting loop along its axis with its south pole entering the loop first. Draw the graphs of

1. The induced current,
2. Joule heating as a function of time. Take the induced current to be positive, if it is clockwise when viewed along the path of the magnet.

Answer:

1. When the south pole of the magnet moves towards the conducting loop along its axis, then according to Lenz’s law, the induced current will be clockwise when viewed along the motion of the magnet.

According to the given question, induced current is positive in this case. Again, when the magnet goes to the other side of the loop.

The direction of the induced current will be anticlockwise. In this case, the induced current is negative. Variations of induced current with time are depicted.

2. Joule heat produced is directly proportional to the square of the current. Hence, the heat produced will always be positive and will be independent of the direction of the current. Variations of heat produced with time are depicted.

Question 12. A semicircular wire of radius r Is rotating with angular velocity ω in a uniform magnetic field B with its radius as the axis. If the resistance of the circuit is R and if the axis of rotation remains perpendicular to B, f what will be the average power produced in each period?

Answer:

Magnetic flux, \(\phi=B A \cos \omega t=B \cdot \frac{\pi r^2}{2} \cdot \cos \omega t\)

∴ emf induced, \(e=-\frac{d \phi}{d t}=\frac{B \pi r^2}{2} \omega \sin \omega t\)

Power, \(P=\frac{e^2}{R}=\frac{\left(B \pi r^2 \omega\right)^2}{4 R} \sin ^2 \omega t\)

In a complete period, the average of \(\sin ^2 \omega t=\frac{1}{2}\)

So, the average power, \(\bar{P}=\frac{\left(B \pi r^2 \omega\right)^2}{8 R}\)

Question 13. Write down the dimensional formula for induced emf.

Answer: Induced emf e has the dimension of potential difference V.

Potential difference \((V)=\frac{work (W)}{charge (Q)}\)

∴ \([e]=[V]=\frac{[W]}{[Q]}=\frac{M^2 T^{-2}}{I T}=\left.M^2 T^{-3}\right|^{-1}\).

Question 14. A plot of magnetic flux (Φ) versus current (I) is shown In the figure for two Inductors A and B. Which of the two has a larger value of self-inductance?

Answer:

As we know,

magnetic flux (Φ) = self-inductance (L) x current(I)

Or, \(L=\frac{\phi}{I}\) = slope of the graph

The slope of the Φ-I graph for inductor A is greater than that for inductor B. Therefore inductor A has a larger value of self-inductance.

Practice Questions on Induced EMF

Question 15. A circular conducting R is placed with its plane perpendicular to a magnetic field. The magnetic field varies with time according to the equation B = B₀sin ωt. Obtain the expression for the Induced current in the coll. coll of radius a and resistance

Answer:

As we know, induced current,

⇒ \(I=\frac{\text { induced emf }(e)}{\text { resistance }(R)}\)

Again, \(e=-\frac{d \phi}{d t}\) = rate of change of magnetic flux with time

∴ \(I=\frac{-\frac{d \phi}{d t}}{R}=-\frac{1}{R} \cdot \frac{d}{d t}\left(B A \cos 0^{\circ}\right)\)

Or, \(I=-\frac{A}{R} \cdot \frac{d}{d t}\left(B_0 \sin \omega t\right)\) [∵B = B₀ sin ω t]

⇒ \(-\frac{A \omega B_0}{R} \cos \omega t=-\frac{\pi a^2 \omega B_0}{R} \cos \omega t\)

Question 16. A conducting wire is bent in the form of an angle θ. The wire moves with velocity v along the bisector of ∠A ‘OB’ as shown. Find the EMF induced between the two ends of the wire If a magnetic field Is acting perpendicular to the plane of the paper and directed into It.

Answer:

Let, 21 = length of the wire A’OB’

where, OA’ = OB’ =l

∴ \(v l \sin \frac{\theta}{2}\) = area described in unit time by the portion OA’.

⇒ \(B l v \sin \frac{\theta}{2}\) = magnetic flux intercepted by OA’ in unit time

= emf induced between O and A’.

Similarly, \(B l v \sin \frac{\theta}{2}\) = emf induced between O and B’.

Using Flemming’s right-hand rule, we get that the induced emf is directed from B’ to A’, i.e., A’ is at a higher potential.

∴ Net induced emf = \(2 \times B l v \sin \frac{\theta}{2}=2 B l v \sin \frac{\theta}{2}\).

Question 17. A Conducting Rod ab of length l, mass m and resistance R slides on a smooth, thick pair of metallic rails. The plane of the rails makes an angle of θ with the Horizontal. A magnetic field B acts along the perpendicular to the horizontal plane in an upward direction. If the rod slides down on the rails at a constant speed v, then shown that \(B=\sqrt{\frac{m g R \sin \theta}{i^2 v \cos ^2 \theta}}\)

Answer:

During the sliding of the rod with constant speed, the net force acting on the rod is zero.

∴ F_mcosθ = mgsinθ

or, BIlcosθ = mgsinθ

or, \(m g \sin \theta=B\left(\frac{B l v \cos \theta}{R}\right) l \cos \theta\)

or, \(m g \sin \theta=\frac{B^2 l^2 v \cos ^2 \theta}{R}\)

or, \(B^2=\frac{m g R \sin \theta}{l^2 v \cos ^2 \theta}\)

∴ \(B=\sqrt{\frac{m g R \sin \theta}{l^2 v \cos ^2 \theta}}\) (Proved)