38: Electromagnetic Induction : Neet-xii-physics

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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 3

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#
Section : iii

Qstn #1
A bar magnet is moved along the axis of a copper ring placed far away from the magnet. Looking from the side of the magnet, an anticlockwise current is found to be induced in the ring. Which of the following may be true?
(a) The south pole faces the ring and the magnet moves towards it.
(b) The north pole faces the ring and the magnet moves towards it.
(c) The south pole faces the ring and the magnet moves away from it.
(d) The north pole faces the ring and the magnet moves away from it.
digAnsr: b,c
Ans : (b) The north pole faces the ring and the magnet moves towards it.
(c) The south pole faces the ring and the magnet moves away from it.
It can be observed that the induced current is in anti-clockwise direction. So, the magnetic field induced in the copper ring is towards the observer.
According to Lenz's law, the current induced in a circuit due to a change in the magnetic flux is in such direction so as to oppose the change in flux.
Two cases are possible:
(1) The magnetic flux is increasing in the direction from the observer to the circular coil.
(2) The magnetic flux is decreasing in the direction from the coil to the observer.
So, from the above mentioned points, the following conclusions can be made:
1. The south pole faces the ring and the magnet moves away from it.
2. The north pole faces the ring and the magnet moves towards it.
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Qstn #2
A conducting rod is moved with a constant velocity v in a magnetic field. A potential difference appears across the two ends
(a) if
v →∥l →
(b) if
v →∥B →
(c) if
l →∥B →
(d) none of these.
digAnsr: d,i,ii,iii
Ans : (d) none of these
The potential difference across the two ends is given by
`` e=Bvl``
It is non-zero only
(i) if the rod is moving in the direction perpendicular to the magnetic field (`` \stackrel{\to }{v}\perp \stackrel{\to }{B}``)
(ii) if the velocity of the rod is in the direction perpendicular to the length of the rod
( `` \stackrel{\to }{v}\perp \stackrel{\to }{l}``)
(iii) if the magnetic field is perpendicular to the length of the rod `` \stackrel{\to }{l}\perp \stackrel{\to }{B}``
Thus, none of the above conditions is satisfied in the alternatives given.
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Qstn #3
A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An emf is induced in the loop if
(a) it is translated
(b) it is rotated about its axis
(c) it is rotated about a diameter
(d) it is deformed.
digAnsr: c,d
Ans : (c) it is rotated about a diameter
(d) it is deformed
When translated or rotated about its axis, the magnetic flux through the loop does not change; hence, no emf is induced in the loop.
When rotated about a diameter, the magnetic flux through the loop changes and emf is induced. On deforming the loop, the area of the loop inside the magnetic field changes, thereby changing the magnetic flux. Due to the change in the flux, emf is induced in the loop.
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Qstn #4
A metal sheet is placed in front of a strong magnetic pole. A force is needed to
(a) hold the sheet there if the metal is magnetic
(b) hold the sheet there if the metal is nonmagnetic
(c) move the sheet away from the pole with uniform velocity if the metal is magnetic
(d) move the sheet away from the pole with uniform velocity if the metal is nonmagnetic.
Neglect any effect of paramagnetism, diamagnetism and gravity.
digAnsr: a,c,d
Ans : (a) hold the sheet there if the metal is magnetic
(c) move the sheet away from the pole with uniform velocity if the metal is magnetic
(d) move the sheet away from the pole with uniform velocity if the metal is nonmagnetic
The strong magnetic pole will attract the magnet, so a force is needed to hold the sheet there if the metal is magnetic.
If we move the metal sheet (magnetic or nonmagnetic) away from the pole, eddy currents are induced in the sheet. Because of eddy currents, thermal energy is produced in it. This energy comes at the cost of the kinetic energy of the plate; thus, the plate slows down. So, a force is needed to move the sheet away from the pole with uniform velocity.
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Qstn #5
A constant current i is maintained in a solenoid. Which of the following quantities will increase if an iron rod is inserted in the solenoid along its axis?
(a) magnetic field at the centre
(b) magnetic flux linked with the solenoid
(c) self-inductance of the solenoid
(d) rate of Joule heating.
digAnsr: a,b,c
Ans : (a) magnetic field at the centre
(b) magnetic flux linked with the solenoid
(c) self-inductance of the solenoid
Iron rod has high permeability. When it is inserted inside a solenoid the magnetic field inside the solenoid increases. As magnetic field increases inside the solenoid thus the magnetic flux also increases. The Self-inductance (L) of the coil is directly proportional to the permeability of the material inside the solenoid. As the permeability inside the coil increases. Therefore, the self-inductance will also increase.
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Qstn #6
Two solenoids have identical geometrical construction but one is made of thick wire and the other of thin wire. Which of the following quantities are different for the two solenoids?
(a) self-inductance
(b) rate of Joule heating if the same current goes through them
(c) magnetic field energy if the same current goes through them
(d) time constant if one solenoid is connected to one battery and the other is connected to another battery.
digAnsr: b,d
Ans : (b) rate of Joule heating if the same current goes through them
(d) time constant if one solenoid is connected to one battery and the other is connected to another battery
Because the solenoids are identical, their self-inductance will be the same.
Resistance of a wire is given by
`` R=\rho \frac{l}{A}``
Here,
l = Length of the wire
A = Area of cross section of the wire
ρ = Resistivity of the wire
Because ρ and l are the same for both wires, the thick wire will have greater area of cross section and hence less resistance than the thin wire.
`` \Rightarrow {R}_{thick}<{R}_{thin}``
The time constant for a solenoid is given by
`` \tau =\frac{L}{R}``
`` \therefore {\tau }_{thick}>{\tau }_{thin}``
Thus, time constants of the solenoids would be different if one solenoid is connected to one battery and the other is connected to another battery.
Also, because the self-inductance of the solenoids is the same and the same current flows through them, the magnetic field energy given by `` \frac{1}{2}L{i}^{2}`` will be the same.
Power dissipated as heat is given by
`` P={i}^{2}R``
i is the same for both solenoids.
`` \therefore {P}_{thick}<{P}_{thin}``
Because the resistance of the coils are different, the rate of Joule heating will be different for the coils if the same current goes through them.
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Qstn #7
An LR circuit with a battery is connected at t = 0. Which of the following quantities is not zero just after the connection?
(a) Current in the circuit
(b) Magnetic field energy in the inductor
(c) Power delivered by the battery
(d) Emf induced in the inductor
digAnsr: d
Ans : (d) Emf induced in the inductor
At time t = 0, the current in the L-R circuit is zero. The magnetic field energy is given by
`` U=\frac{1}{2}L{i}^{2}``, as the current is zero the magnetic field energy will also be zero. Thus, the power delivered by the battery will also be zero. As, the LR circuit is connected to the battery at t = 0, at this time the current is on the verge to start growing in the circuit. So, there will be an induced emf in the inductor at the same time to oppose this growing current.
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Qstn #8
A rod AB moves with a uniform velocity v in a uniform magnetic field as shown in figure.
(a) The rod becomes electrically charged.
(b) The end A becomes positively charged.
(c) The end B becomes positively charged.
(d) The rod becomes hot because of Joule heating.
Figure
digAnsr: b
Ans : (b) The end A becomes positively charged.
Due to electromagnetic induction, emf e is induced across the ends of the rod. This induced emf is given by
`` e=Bvl``
The direction of this induced emf is from A to B, that is, A is at the higher potential and B is at the lower potential. This is because the magnetic field exerts a force equal to `` qvB`` on each free electron where q is `` -``1.6 × 10^-16 C. The force is towards AB by Fleming's left-hand rule; hence, negatively charged electrons move towards the end B and get accumulated near it. So, a negative charge appears at B and a positive charge appears at A.
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Qstn #9
L, C and R represent the physical quantities inductance, capacitance and resistance respectively. Which of the following combinations have dimensions of frequency?
(a)
1RC
(b)
RL
(c)
1LC
(d) C/L.
digAnsr: a,b,c
Ans : (a) `` \frac{1}{RC}``
(b) `` \frac{R}{L}``
(c) `` \frac{1}{\sqrt{LC}}``
The time constant of the RC circuit is given by
`` \tau =RC``
On taking the reciprocal of the above relation, we get
`` {f}_{1}=\frac{1}{RC}`` ...(1)
f₁ will have the dimensions of the frequency.
The time constant of the LR circuit is given by
`` \tau =\frac{L}{R}``
On taking the reciprocal of the above relation, we get
`` {f}_{2}=\frac{R}{L}`` ...(2)
f₂ will have the dimensions of the frequency.
On multiplying eq. (1) and (2), we get
`` {f}_{1}{f}_{2}=\frac{1}{LC}``
`` \Rightarrow \sqrt{{f}_{1}{f}_{2}}=\frac{1}{\sqrt{LC}}``
Thus, `` \sqrt{{f}_{1}{f}_{2}}`` will have the dimensions of the frequency.
Page No 305:

Qstn #10
The switches in figure
(a) and
(b) are closed at t = 0 and reopened after a long time at t = t₀.
Figure
(a) The charge on C just after t = 0 is εC.
(b) The charge on C long after t = 0 is εC.
(c) The current in L just before t = t₀ is ε/R.
(d) The current in L long after t = t₀ is ε/R.
digAnsr: b,d
Ans : (b) The charge on C long after t = 0 is εC.
(d) The current in L long after t = t₀ is ε/R.
The charge on the capacitor at time ''t'' after connecting it with a battery is given by,
`` Q=C\epsilon \left[1-{e}^{-t/RC}\right]``
Just after t = 0, the charge on the capacitor will be
`` Q=C\epsilon \left[1-{e}^{0}\right]=0``
For a long after time, `` t\to \infty ``
Thus, the charge on the capacitor will be
`` Q=C\epsilon \left[1-{e}^{-\infty }\right]``
`` \Rightarrow Q=C\epsilon \left[1-0\right]=C\epsilon ``
The current in the inductor at time ''t'' after closing the switch is given by
`` I=\frac{{V}_{\,\mathrm{\,b\,}}}{R}\left(1-{e}^{-tR/L}\right)``
Just before the time t₀, current through the inductor is given by
`` I=\frac{{V}_{\,\mathrm{\,b\,}}}{R}\left(1-{e}^{-{t}_{0}R/L}\right)``
It is given that the time t₀ is very long.
∴ `` {t}_{0}\to \infty ``
`` I=\frac{\epsilon }{R}\left(1-{e}^{-\infty }\right)=\frac{\epsilon }{R}``
When the switch is opened, the current through the inductor after a long time will become zero.
Page No 306:

#
Section : iv

Qstn #1
Calculate the dimensions of

#1-a
∫E→.dl,→
Ans : The quantity`` \int ``E.dl can also be written as:
`` \int ``E.dl = V (V = Voltage)
Unit of voltage is J/C.
Voltage can be written as:
`` \,\mathrm{\,Voltage\,}=\frac{\,\mathrm{\,Energy\,}}{\,\mathrm{\,Charge\,}}``
Dimensions of energy = [ML²T^-2]
Dimensions of charge = [IT]
Thus, the dimensions of voltage can be written as:
[ML²T^-2] ×[IT]^-1 = [ML²I^-1T^-3]

#1-b
vBl and
Ans : The quantity vBl is the product of quantities v, B and L.
Dimensions of velocity v = [LT^-1]
Dimensions of length l = [L]
The dimensions of magnetic field B can be found using the following formula:
`` B=\frac{F}{qv}``
Dimensions of force F = [MLT^-2^]
Dimensions of charge q = [IT]
Dimensions of velocity = [LT^-1]
The dimensions of a magnetic field can be written as:
MI^-1T^-2
∴ Dimensions of vBl = [LT^-1] × [MI^-1T^-2] × [L]= [ML²I^-1T^-3]