BSEB Class 12 Physics Solutions Short AnswersType Questions Electromagnetic Induction Additional Important Questions

BSEB Class 12 Physics Solutions Short Answers Type Questions Electromagnetic Induction

Class 12 Physics Chapter 6 Electromagnetic Induction  Short Answers Additional Important Questions Type Questions

Bihar Board Class 12 Physics Solutions Chapter 6 Electromagnetic Induction Textbook Questions and Answers, Additional Important Questions, Notes.

Short Answer Type Questions BSEB Class 12 Physics Solutions Short Answers

Question 1.
State Faraday’s laws of E.M.I.
Answer:
The following are the three laws of E.M.I. given by Faraday.
(i) An induced e.m.f. is always produced in a circuit or a coil whenever there is a change of magnetic flux linked with it.
(ii) The induced e.m.f. lasts so long as the change in the magnetic flux continues.
(iii) The magnitude of the induced e.m.f. is directly proportional to the rate of change of the magnetic flux linked with the circuit.
i.e., mathematically, e ∝ \frac{\phi_{2}-\phi_{1}}{\mathrm{t}}
or e = -k (\frac{\phi_{2}-\phi_{1}}{t})
where k is proportionality constant.
ø1 ø2are initial and final values of magnetic flux linked with the coil.
t = time in which the flux changes from to ø1 to ø2 Negative sign shows that e is of opposing nature. In S.l. system, k = I
e = \frac{\phi_{2}-\phi_{1}}{t}
or e = – \frac {dø}{dt}
If the coil has N turns, and Ø be the flux of each turn, then
e = – \frac {d}{dt} (Nø) = – N\frac {dø}{dt} .

Question 2.
Prove that Lenz’s law is in accordance with the law of conservation of energy.
Answer:
In the electromagnetic induction, the electrical energy (in the form of induced current or induced e.m.f.) is produced at the cost of mechanical energy. To varify it, let us consider a closed coil connected to a galvanometer. Let a bar magnet faces with its N-pole towards coil. Now move the magnet towards the coil. The magnetic flux linked with it will go on increasing, thus induced e.m.f. is produced in the coil. Due to this, induced current flows through the galvanometer which shows the deflection. According to Lenz’s law, the direction of induced e.m.f. is such that it opposes the cause z.e., motion of magnet which has produced it. To oppose the cause, the upper face of the coil acquires N-polarity, thus repels the magnet. To move the magnet towards the coil, mechanical work has to be done to overcome the force of repulsion between the N-poles. This work done is converted into electrical energy.

S2

Similarly when the magnet moves away from the coil, it face near the magnet acquires S-polarity so as to oppose its motion. Again the work has to be done to overcome the force of attraction between S and N-pole of the coil and magnet respectively and the work is converted into electrical energy. Hence Lenz’s Law is in accordance with the law of conservation of energy.

BSEB Class 12 Physics Solutions Short Answers

Question 3.
Desire the expression for the motional e.m.f.
Answer:
Consider a closed circuit ABCD. Let a uniform magnetic field
\overrightarrow{\mathrm{B}} be directed into the plane of paper and this circuit is placed in \overrightarrow{\mathrm{B}}
Let \overrightarrow{\mathrm{V}} be the velocity of the circuit towards right. If \overrightarrow{\mathrm{f}_{\mathrm{m}}} be the magnetic Lorentz force experienced by each electron inside the wire, then
\overrightarrow{\mathrm{f}_{\mathrm{m}}} = -e (\overrightarrow{\mathrm{V}} x \overrightarrow{\mathrm{B}}) ……..(1)
where \overrightarrow{\mathrm{f}_{\mathrm{m}}} = – e charge on an electron and acts along BC.

S3S3

Due to this force the electrons accumulate at end C, thus make C negatively charged while B is positively charged. Hence an electric field \overrightarrow{\mathrm{E}} is set up between the two ends of the wire BC along B to C, so the electrons experience an electrostatic force \overrightarrow{\mathrm{f}_{\mathrm{e}}} = – e \overrightarrow{\mathrm{E}} along C to B under equilibrium condition, the net force on an electron becomes zero i.e.,

S3 a

If ε be the induced e.m.f. produced in arm BC of length l, then
E = \frac {ε}{l} ………(4)
From (3) and (4), we get
\frac {ε}{l} = υB
or ε = υlB which is the required expression for the motional e.m.f.

Aliter :
Let dø be the decrease in the magnetic flux of the circuit ABCD when it moves out of magnetic field by a small distance dx in a time dt. ‘
dø = – B. x BC x dx = – B, l dx, where l = length
If ε be the induced e.m.f. produced in the circuit, then according to Faraday’s rule,

S3 b

If R be the resistance of the closed circuit, then the induced current flowing through the circuit is given by
I = \frac {E}{R} = \frac {Blυ}{R}

Question 4.
Derive the expression for the force required to pull a conductor or a rod out of magnetic field.
Answer:
Let υ be the velocity of the conductor of kngth I moving through a constant magnetic field \overrightarrow{\mathrm{B}} .
Let I = Current flowing through the conductor.
If E be the motional e.m.f. produced in the conductor, then
I = \frac {ε}{R} = \frac {Blυ}{R} ……..(1)
Due to this current in the conductor, it experiences a force. To pull the conductor with a constant speed z’, this force must be balanced by the pulling force.
Let F be the pulling force.
∴ Fυ = mechanical power supplied to the circuit or conductor …(2)
Also electric power = rate at which electric energy is supplied = rate at which electric work s done.
= ε.l = \frac{\varepsilon^{2}}{\mathrm{R}}=\frac{(\mathrm{B} l v)^{2}}{\mathrm{R}} ……..(3)
∴From (2) and (3), we get
Fυ = \frac{\mathrm{B}^{2} l^{2} v^{2}}{\mathrm{R}}
or F = \frac{\mathrm{B}^{2} l^{2} v}{\mathrm{R}}
which is the required expression for thL pulling force.

Question 5.
What are the advantages or application of eddy currents?
Answer:
The following are the advantages of eddv currents.
(i) Diathermy : These are used for deep heat treatment i.e., localised heating of tissues in human body. This treatment is called Diathermy.
(ii) Concept of eddy currents is used in energy metres to record the consumption of electricity.
(iii) Speedometer : The eddy currents are produced in the aluminium drum in which a magnet attached to the axle of the wheel rotates according to the speed of the vehicle. The pointer attached to the drum is deflected in the direction of rotation of the drum.
(iv) Electromagnetic brakes : These are used in trains. The large eddy currents are produced in the metallic drum which oppose the motion of the drum.
(v) Dead beat galvanometer : Electromagnetic damping is used to stop coils in a shorter interval of time. When the coil wound over a metallic frame is deflected, eddy currents are produced ‘ in the metallic frame which oppose the motion of the coil.
(vi) Induction Furnace makes use of heating effect of edd v currents.
(vii) Induction motor or a.c. motor also works bv eddv currents.

BSEB Class 12 Physics Solutions Short Answers

Question 6.
Prove that e = – L \frac {dI}{dt} where the symbols have their usual meanings.
Answer:
Let I be the current flowing through a coil. Also let (j) be the magnetic flux linked with the coil at any instant. It is found that at any instant,
ø ∝ I.
or ø ∝ LI ……(i)
Where L is the proportionality constant known as coefficient of self-induction. If s be the induced e.m.f. produced in the coil, then
ε = – \frac {dø}{dt} = –\frac {d}{dt}(LI) = -L\frac {dI}{dt}.

Question 7.
Prove that c or e = M where the symbols have their usual meaning.
Answer:
Let P and S be the two coils as shown in figure. If I be the current flowing through the P coil at any instant, then the magnetic flux linked with the S-coil at that instant is found to be directly proportional to I i.e.,
ø ∝ I
ø ∝ MI
where M is the coefficient of mutual induction or simply mutual- induCtance of two coils.

S7

If s be the induced e.m.f. produced in the S-coil, then
ε = – \frac {dø}{dt} = –\frac {d}{dt}(MI) = –\frac {dI}{dt}.

BSEB Class 12 Physics Solutions Short Answers

Question 8.
On what factors does the coefficients of mutual inductance depends?
Answer:
It depends upon the following factors:
(i) Total number of turns in the two coils.
(ii) Nature of the material on which the two coils are wound.
(iii) Size of the coils.
(iv) Distance between two coils.
(v) Relative placement of two coils i.e., their orientation.

Question 9.
On what factors does the coefficient of self-inductance of a given coil depends?
Answer:
It depends upon:
(i) The number of turns in the coil.
(ii) area of cross-section of the coil. More the area of the conductor, lesser will be its resistance and thus more is the rate of change current and thus lesser will be L.
(iii) the permeability of the medium over which the col is wound.
(iv) the length of the coil.

Question 10.
Derive the expression for self-inductance of a long-solenoid?
Answer:
A solenoid is said to be long if its length is very large as compared to its radius of the cross-section.
Let N = total no. of turns in the solenoid of length l and area of cross-section A.
If n be the no. of turns per unit length of the solenoid, then
n = \frac {N}{l} ……….(i)
Let I be the current flowing through the solenoid. If B be the magnetic field at a point inside the solenoid, then
B = μ0nl ……..(ii)
which remains constant for a long solenoid.
Let ø1 be the magnetic flux linked with each turn of the solenoid.
ø1 = B x A = μ0nllA. .
If ø be the total magnetic flux linked with the solenoid, then
ø = N x ø1 = nlø1 = nl x μ0 nlA
= μ0 n2IAl = \frac{\mu_{0} \mathrm{~N}^{2} \mathrm{IA}}{l}
Also ø = LI
∴ L = μ0 N2IA/l

BSEB Class 12 Physics Solutions Short Answers

Question 11.
Derive the expression for the induced charge flowing in a coil in terms of: (a) magnetic flux, (b) magnetic field.
Answer:
Let R be the resistance of a coil.
I = induced current in it.
If e be the induced e.m.f. produced in the coil, then according to Faraday’s law,
e = – \frac {dø}{dt} …(1)
Let q be the total charge induced in the coil. If dq be the charge induced in a small time dt, then

S11

Let ø1 and ø2 be the initial and final values of the magnetic flux linked with the coil when charge on it changes from 0 to q. Then integrating (2), we get

S11 a 1

or magnitude of q is given by

(b) Let N= total no. of turns in the coil.
A = area of the coil.
B rragnetic field in vhR h the coi) ís pla4ed

If ø be the total magnetic flux linked with the coil, then
ø = NBA ……..(4)
Also from(2), dq = – \frac {dø}{R}
integrating, \int_{0}^{q} \mathrm{~d} q=\int_{0}^{\phi} \frac{(-\mathrm{d} \phi)}{\mathrm{R}}
or q =- \frac{1}{\mathbf{R}}[\phi]_{0}^{\phi}=-\frac{\phi}{\mathrm{R}}
or magnitude q is given by
q = \frac{\phi}{R}=\frac{\text { NBA }}{R}

Question 12.
What is the physical significance of self-inductance in an electrical circuit?
Answer:
It plays the same role in electrical circuit as is being played bv mass or inertia in mechanical motion (i.e., translatory and rotatory motions respectively.)

Question 13.
Derive the expression for the energy stored in a solenoid having self-inductance L.
Answer:
Let us consider a source of e.m.f. connected to a coil of inductance L. As the current in it starts growing, an induced e.m.f. is set up in the inductor which opposes the growth of the current in it. The source of e.m.f. has to spend energy in sending the current through the circuit against the induced e.m.f. which is stored in the inductor in the
form of induced e.m.f. initally the current in the induction is zero. Let \frac {dl}{dt} be the rate of growth of current in the inductor at the instant when a current l flows through it. If e be the induced e.m.f. produced in the coil, then
e = L \frac {dl}{dt} ……..(i)
Let dW be the work done by the source in sending the current 1 for a time dt, then
dW = eldt = L\frac {dl}{dt} Idt = IL dl
Let dW be the total work done in increasing the current from 0 to I0 then

S13

which is stored in the inductor in the form of energy U,
U = \frac {dl}{dt} LI02.

Question 14.
What you conclude from U = \frac {dl}{dt} LI2.
Answer:
From this expression, we conclude that:
(i) The energy is stored in the inductor at the expense of energy of the source of e.m.f.
(ii) It resides in the inductor in the form of magnetic field.
(iii) The average electric power of an inductor is zero.

BSEB Class 12 Physics Solutions Short Answers

Question 15.
Derive the expression for the mutual inductance of two long solenoids.
Answer:
Let S1 and S2 be the two long solenoids of same length / such that S2 closely wound over S1 so that both have nearly same area of cross-section A.

S15 1

Lei n1 and n2 be the no. of turns per unit length in S1 and S2 respectively.
If B1 be the magnetic field produced at any point inside S1 due to the current I1 in it, then
B1 = µ0n1I1 ……(1)
Now let ø2 be the magnetic flux linked with each turn of the solenoid.
ø2 = B1 x area of each turn = B1A …..(ii)
If ø be the total magnetic flux linked with S2, then
ø = ø2 x Total number of turns in S2.
= ø2 x n2x l = B1 A x n2l
= ø2n1l1An2l = µ0n1n2I1lA …….(iii)
Also ø ∝ F1 or ø = MI1 …….(iv)
∴ From (iii) and (iv), we get
MI1 = µ0n1n2lAI1
or M = µ0n1n2lA ……..(v)
If N1 and N2be the total no. of turns in each coil, then N1 = n1l and N2 = n2l.
∴ (v) reduces to
M = µ0 \frac{N_{1}}{l} \cdot \frac{N_{2}}{l} Al
M = µ0 \frac{\mathbf{N}_{1} \mathbf{N}_{2} \mathbf{A}}{1}

Question 16.
What will happen if there is a relative motion between the magnet and the open coil?
Answer:
In case of a coil having open ends, the induced e.m.f. is set up across the open ends of the coil due to flow of induced current when there is a relative motion between the magnet and the open coil.

Question 17.
Prove that induced current in a coil is given by I ≈ \frac {N}{R} A (B1 – B1) where the symbols have their usual mean.
Answer:
We know from Q. 11 that
q = – \frac {1}{R}2 – ø1)
Or L.dt = \frac {1}{R} dø …….(1)
Where q = I dt = induced charged produced in the coil in a fine dt when a current I flows through it.
dø = ø2 – ø1 = change in magnetic flux.
∴ I = – \frac {1}{R} \frac {dø}{dt} ……(2)
When the magnetic field is parallel to the outward drawn normal to the surface of the coil, then θ = 0°.
∴ ø1 = B1 A and ø2 = B2A.
dø = (B2 – B1)A. ……(3)
If N be the total no. of turns in the coil, then ‘
dø = N (B2 – B1) A ……(4)
∴ From (2) and (4), we get
I = – \frac {1}{R} N(B2 – B1) A
I = \frac {NA}{R} (B2 – B1)

Question 18.
Explain why the word generator is a misnomer? Define A.C. generator.
Answer:
A.C. generator simply converts mechanical energy into electrical energy. Thus it would be better to call it a converter. It is a device which converts stored chemical energy into alternating electrical energy.

Question 19.
Explain the working of A.C. generator.
Answer:
As per Q1 (long answer type Question), the e.mi. induced by rotation of a coil in a magnetic field is given by
ε = NBA ω sin ωt = ε0,sin ωt
where ε0 = NBAco is the amplitude of e.m.f. produced in the rotating coil.
If Rbe the resistance of the circuit, then the instantaneous current produced in the circuit is given by :

BSEB Class 12 Physics Solutions Short Anser

when I0 = \frac{\varepsilon_{0}}{\mathrm{R}} is the maximum value of instantaneous current.

Class 10 Objective Click Here

12th Physics Objective Question स्थिरवैद्युत विभव तथा धारिता MCQ

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