(a) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it.
(b) The current flowing through an inductor of self-inductance L is continuously increasing. Plot a graph showing the variation of
(i) Magnetic flux versus the current
(ii) Induced emf versus dI/dt
(iii) Magnetic potential energy stored versus the current.
(a) The statement ‘polarity of the induced emf is such that it opposes a change in magnetic flux’ is given by Lenz law.
The given activity demonstrates the above statement. The amount of magnetic flux linked with the coil increases, when the north pole of a bar magnet is brought near the coil. Current in the coil is induced in a so as to opposes the increase in magnetic flux. This is possible only when the current induced in the coil is in anti-clockwise direction, with respect to an observer. The magnetic moment associated with this induced emf has north polarity, towards the north pole of the approaching bar magnet.
Similarly, magnetic flux linked with the coil decreases when the north pole of the bar magnet is moved away from the coil. Inorder to oppose this decrease in magnetic flux, current is induced in the coil in clockwise direction so that its south pole faces the receding north pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in magnetic flux.
(b) (i) Since ϕ = L I ;
where,
I = Strength of current through the coil at any time,
ϕ = Amount of magnetic flux linked with all turns of the coil at that time, and
L = Constant of proportionality called coefficient of self-induction.
(ii) Induced emf, e=− = 
i.e., e = − 
(iii) Since magnetic potential energy is given by , 
State the working of a.c. generator with the help of a labelled diagram.
The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity. Deduce the expression for the alternating e.m.f. generated in the coil.
What is the source of energy generation in this device?
OR
(a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by 
/2 in phase.
(b) A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
(i) Write the expression for the instantaneous value of the e.m.f. induced in the wire.
(ii) What is the direction of the e.m.f.?
(iii) Which end of the wire is at the higher potential?
Working of ac generator:
When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil, its direction being given by Fleming’s right hand rule. Considering the armature to be in vertical position and as it rotates in anticlockwise direction, the wire ab moves upward and cd downward, so that the direction of induced current is shown in fig. In the external circuit, the current flows along B1RL B2.
The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves downward and cd upward, so the direction of current is reversed and in external circuit it flows along B2RL B1. Thus the direction of induced emf and current changes in the external circuit after each half revolution.
Let, N be the number of turns in the coil,
f is the frequency of rotation,
A is the area of coil,
B is the magnetic induction.
Then, the induced emf is given by, 
We can see that the emf and current produced is alternating.
Current produced by an ac generator cannot be measured by moving coil ammeter; because the average value of ac over full cycle is zero.
The source of energy generation is the mechanical energy of rotation of armature coil.
OR
Consider a coil of self-inductance L and negligible ohmic resistance. An alternating potential difference is applied across the ends. The magnitude and direction of AC is changing periodically, changing the magnetic flux. So, an emf is induced and produced in the coil. The instantaneous value of alternating voltage applied is given by,
V = Vo sin w t
Instantaneous induced emf is, 
According to Kirchhoff’s second law in closed circuit, we get
Integrating with respect to time ‘t’,
which is the required expression for current.
;
where, 
is the peak value of alternating current.
From the above equations we can say that the current lags behind the applied voltage by an angle .
b)
i) Induced emf, = BHvL ; where BH is horizontal component of earth’s magnetic field directed from South to North.
ii) West to East
iii) East end of the wire is at higher potential
(a) Define mutual inductance and write its S.I. units.
(b) Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other.
(c) In an experiment, two coils C1 and C2 are placed close to each other. Find out the expression for the emf induced in the coil C1 due to a change in the current through the coil C2.
a) Mutual inductance of two coils is equal to the e.m.f induced in one coil when rate of change of current through the other coil is unity.
SI unit of mutual inductance is henry.
b) Consider two long solenoids S1 and S2 of same length ‘l’ such that S2 surrounds S1 completely.
Let,
n1 = Number of turns per unit length of S1
n2 = Number of turns per unit length of S2
I1 = Current passing through solenoid S1
∅21 = Flux linked with S2 due to current flowing in S1
is the coefficient of mutual inductance of two solenoids.
When current is passed through S1, emf is induced in S2 .
Magnetic field inside solenoid S1 is given by,
Total magnetic flux liked with S2 is given by,
Similarly, mutual inductance between two solenoids, when current is passed through S2 and emf induced in solenoid S1 is given by, 
Hence, coefficient of mutual induction between the two long solenoids is given by, 
c) It is found that, 
where, I is the strength of current in coil 2, and ∅ is the total amount of magnetic flux linked with coil 1.
Emf induced in the neighboring coil C1 is,
(i) With the help of a labelled diagram, describe briefly the underlying principle and working of a step up transformer.
(ii) Write any two sources of energy loss in a transformer.
(iii) A step up transformer converts a low input voltage into a high output voltage.
Does it violate law of conservation of energy? Explain.
i) Underlying principle of a step-up transformer:
A transformer which increases the ac voltage is known as a step up transformer.
Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.
i) Underlying principle of a step-up transformer: A transformer which increases the ac voltage is known as a step up transformer.
Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.
Working: When an A.C source is connected to the ends of the primary coil, the current changes continuously in the primary coil. Hence, the magnetic flux which is linked with the secondary coil changes continuously. So, the emf which is developed across the secondary coil is same as that in the primary coil. The emf is induced in the coil as per Faraday’s law.
Assumption: We assume that there is no leakage of flux so that, the flux linked with each turn of primary coil is same as flux linked with secondary coil.
ii) Two sources of energy loss in the transformer:
Joule Heating – Energy is lost in resistance of primary and secondary windings in the form of heat.
H = I2 Rt
Flux leakage - Energy is lost due to coupling of primary and secondary coils not being perfect, i.e., whole of magnetic flux generated in primary coil is not linked with the secondary coil.
iii) Conservation of law of energy is not violated in step-up transformer. When output voltage increases, the output current automatically decreases. Thus, there is no loss of energy.
(a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of variation of
(i) Magnetic flux and
(ii) Alternating emf versus time generated by a loop of wire rotating in a magnetic field.
(b) Why is choke coil needed in the use of fluorescent tubes with ac mains?(a) The working of AC generator is based on the principle of electromagnetic induction.
Construction:
AC generator mainly consists of:
Armature − The rectangular coil ABCD
Filed Magnets − Two pole pieces of a strong electromagnet
Slip Rings − The ends of the coil ABCD are connected to two hollow metallic rings R1 and R2.
Brushes − B1 and B2 are two flexible metal plates or carbon rods. They are fixed and are kept in tight contact with R1 and R2, respectively.
Working –
The angle θ between field and the normal to the coil changes continuously as the armature coil is rotated in a magnetic field. Therefore, magnetic flux linked with the coil changes and an emf is induced in the coil. According to Fleming’s right hand rule, current is induced from A to B in AB and from C to D in CD. In the external circuit, current flows from B2 to B1.
Magnetic flux linked with the coil is given by,
... (1)
Graph between magnetic flux and time, according to equation (i), is shown below:
As the coil rotates, angle θ changes. Therefore, magnetic flux Φ linked with the coil changes and an emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then
The graph between alternating emf vs. time is shown below:
(b) A choke coil enables us to control the current in an ac circuit. If resistance R is used, then a lot of energy will be wasted in the form of heat.
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