NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction: In the chapter moving charges and magnetism you have studied that moving charges produces a magnetic field. The solutions of NCERT class 12 physics chapter 6 electromagnetic induction focus on the question based on the concept that the changing magnetic field can produce an emf across an electrical conductor. The equipment like an electric generator and the transformer work on this principle. In chapter 1 you have learnt about electric flux and in NCERT grade 12 chapter 6 electromagnetic induction you will learn about magnetic flux. The questions explained in the CBSE NCERT solutions for class 12 physics chapter 6 electromagnetic induction are based on the depth of understanding the concept. Many questions based on finding the directions of the magnetic field and current are explained in the NCERT solutions for class 12 physics chapter 6 electromagnetic induction which will be useful in exams. NCERT solutions are an important tool to perform better in the exams.
Q 6.1(a) Predict the direction of induced current in the situations described by the following
To oppose the magnetic field current should flow in anti-clockwise, so the direction of the induced current is qrpq
Current in the wire in a way such that it opposes the change in flux through the loop. Here hence current will induce in the direction of p--->r--->q in the first coil and y--->z--->x in the second coil.
Q 6.1 (c) Predict the direction of induced current in the situations described by the following Figs.(c)
When we close the key, the current will flow through the first loop and suddenly magnetic flux will flow through it such that magnetic rays will go from right to left of the first loop. Now, to oppose this change currently in the second loop will flow such that magnetic rays go from left to right which is the direction yzxy
Q 6.1 (d) Predict the direction of induced current in the situations described
by the following Fig. (d)
hen we increase the resistance of the rheostat, the current will decrease which means flux will decrease so current will be induced to increase the flux through it. Flux will increase if current flows in xyzx.
On the other hand, if we decrease the resistance that will increase the current which means flux will be an increase, so current will induce to reduce the flux. Flux will be reduced if current goes in direction zyxz
Q 6.1 (e) Predict the direction of induced current in the situations described by the following Fig(e)
As we release the tapping key current will induce to increase the flux. Flux will increase when current flows in direction xryx.
The current will not induce as the magnetic field line are parallel to the plane. In other words, since flux through the loop is constant (zero in fact), there won't be any induction of the current.
Q6.2 (a) Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19: a
A wire of irregular shape turning into a circular shape;
By turning the wire from irregular shape to circle, we are increasing the area of the loop so flux will increase so current will induce in such a way that reduces the flux through it. By right-hand thumb rule direction of current is adcba.
A circular loop being deformed into a narrow straight wire.
Here, by changing shape, we are decreasing the area or decreasing the flux, so the current will induce in a manner such that it increases the flux. Since the magnetic field is coming out of the plane, the direction of the current will be adcba.
Q6.3 A long solenoid with turns per cm has a small loop of area placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from to in , what is the induced emf in the loop while the current is changing?
Given in a solenoid,
The number of turn per unit length :
loop area :
Current in the solenoid :
change in current :
change in time:
Now, the induced emf :
hence induced emf in the loop is .
Given:
Length of rectangular loop :
Width of the rectangular loop:
Area of the rectangular loop:
Strength of the magnetic field
The velocity of the loop :
Now,
a) Induced emf in long side wire of rectangle:
this emf will be induced till the loop gets out of the magnetic field, so
time for which emf will induce :
Hence a emf will be induced for 2 seconds.
b) Induced emf when we move along the width of the rectangle:
time for which emf will induce :
Hence a emf will induce for 8 seconds.
Given
length of metallic rod :
Angular frequency of rotation :
Magnetic field (which is uniform)
Velocity: here velocity at each point of the rod is different. one end of the rod is having zero velocity and another end is having velocity . and hence we take the average velocity of the rod so,
Now,
Induce emf
Hence emf developed is 100V.
Given
The radius of the circular loop
Number of turns
Flux through each turn
Flux through N turn
Induce emf:
Now,
maximum induced emf (when sin function will be maximum)
Average induced emf
as the average value of sin function is zero,
Maximum current when resistance of the loop is .
Power loss :
Here, power is getting lost as emf is induced and emf is inducing because we are MOVING the conductor in the magnetic field. Hence external force through which we are rotating is the source of this power.
Q6.7 (a) A horizontal straight wire long extending from east to west isfalling with a speed of , at right angles to the horizontal component of the earth’s magnetic field, .
What is the instantaneous value of the emf induced in the wire?
Given
Length of the wire
Speed of the wire
The magnetic field of the earth
Now,
The instantaneous value of induced emf :
Hence instantaneous emf induce is .
What is the direction of the emf?
If we apply the Flemings right-hand rule, we see that the direction of induced emf is from west to east.
Which end of the wire is at the higher electrical potential?
The eastern wire will be at the higher potential end.
Given
Initial current
Final current
Change in time
Average emf
Now,
As we know, in an inductor
Hence self-inductance of the circuit is 4H.
Given
Mutual inductance between two coils:
Currents in a coil:
Change in current:
The time taken for the change
The relation between emf and mutual inductance:
Hence, the change in flux in the coil is .
Given
Speed of the plane:
Earth's magnetic field at that location:
The angle of dip that is angle made with horizontal by earth magnetic field:
Length of the wings
Now, Since the only the vertical component of the magnetic field will cut the wings of plane perpendicularly, only those will help in inducing emf.
The vertical component of the earth's magnetic field :
So now, Induce emf :
Hence voltage difference developed between the ends of the wing is 3.125V.
Given,
Area of the rectangular loop which is held still:
The resistance of the loop:
The initial value of the magnetic field :
Rate of decreasing of this magnetic field:
Induced emf in the loop :
Induced Current :
The power dissipated in the loop:
The external force which is responsible for changing the magnetic field is the actual source of this power.
Given,
Side of the square loop
Area of the loop:
The resistance of the loop:
The velocity of the loop in the positive x-direction
The gradient of the magnetic field in the negative x-direction
Rate of decrease of magnetic field intensity
Now, Here emf is being induced by means of both changing magnetic field with time and changing with space. So let us find out emf induced by both changing of space and time, individually.
Induced emf due to field changing with time:
Induced emf due to field changing with space:
Now, Total induced emf :
Total induced current :
Since the flux is decreasing, the induced current will try to increase the flux through the loop along the positive z-direction.
Given,
Area of search coil :
The resistance of coil and galvanometer
The number of turns in the coil:
Charge flowing in the coil
Now.
Induced emf in the search coil
Hence magnetic field strength for the magnet is 0.75T.
Given
Length of the rod
Speed of the rod
Strength of the magnetic field
induced emf in the rod
Hence 9mV emf is induced and it is induced in a way such that P is positive and Q is negative.
(b)Is there an excess charge built up at the ends of the rods when is open? What if is closed?
Yes, there will be excess charge built up at the end of the rod when the key is open. This is because when we move the conductor in a magnetic field, the positive and negative charge particles will experience the force and move into the corners.
When we close the key these charged particles start moving in the closed loop and continuous current starts flowing.
When the key K is open there is excess charge at both ends of the rod. this charged particle creates an electric field between both ends. This electric field exerts electrostatic force in the charged particles which cancel out the force due to magnetic force. That's why net force on a charged particle, in this case, is zero.
Q6.14 (d) Figure 6.20 shows a metal rod resting on the smooth rails and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer connects the rails through a switch . Length of the , , resistance of the closed loop containing the . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of in the direction shown. Give the polarity and magnitude of the induced emf.
(d) What is the retarding force on the rod when K is closed?
Induced emf = 9mV (calculated in a part of this question)
The resistance of loop with rod = 9
Induced Current
Now,'
Force on the rod
Hence retarding force when k is closed is .