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NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction- Welcome to the updated class 12 physics electromagnetic induction NCERT solutions. On this NCERT Solution page, you will find comprehensive and detailed solutions prepared by subject experts from Careers360. These solutions cover a total of seventeen questions, including exercise and additional exercise questions.
Physics class 12 chapter 6 ncert solutions are invaluable for 12th-grade students. This Chapter 6 PDF is available to make the subject more accessible and engaging. To gain a comprehensive understanding of the topic and enhance their Class 12 Physics Chapter 6 Electromagnetic Induction NCERT notes, students should make it a practice to regularly work through these solutions
Hence, students can rely upon the NCERT solutions for Class 12 Physics Chapter 6 Electromagnetic Induction and prepare for the exams. The NCERT exemplar Class 12 Physics Chapter 6 solutions focus on the question based on the concept that the changing magnetic field can produce an emf across an electrical conductor. Electromagnetic induction class 12 ncert solutions are an important tool to perform better in exams. The NCERT solutions for Class 12 also help in the preparations for competitive exams. Read further to know the NCERT solutions for Class 12 Physics Chapter 6 Electromagnetic Induction in detail.
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NCERT solutions for class 12 physics chapter 6 electromagnetic induction: Exercise Question Answer
Q 6.1(a) Predict the direction of induced current in the situations described by the following
Answer:
To oppose the magnetic field current should flow in anti-clockwise, so the direction of the induced current is qrpq
Q 6.1 (b) Predict the direction of induced current in the situations described by the following Figs.
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
As we release the tapping key current will induce to increase the flux. Flux will increase when current flows in direction xryx.
Q 6.1 (f) Predict the direction of induced current in the situations described by the following Fig (f)
Answer:
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;
Answer:
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.
Answer:
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?
Answer:
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 .
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
The eastern wire will be at the higher potential end.
Answer:
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.
Answer:
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 .
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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.
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod even though they do experience magnetic force due to the motion of the rod. Explain.
Answer:
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?
Answer:
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 .
(e) How much power is required (by an external agent) to keep the rod moving at the same speed when is closed? How much power is required when is open?
Answer:
Force on the rod
Speed of the rod
Power required to keep moving the rod at the same speed
Hence required power is 9mW.
When the key is open, no power is required to keep moving rod at the same speed.
Q6.14 (f) 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
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?
Answer:
Current in the circuit = 1A
The resistance of the circuit = 9m
The power which is dissipated as the heat
Hence 9mW of heat is dissipated.
We are moving the rod which induces the current. The external agent through which we are moving our rod is the source of the power.
Q6.14 (g) 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
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Answer:
If the magnetic field is parallel to the rail then, the motion of the rod will not cut across the magnetic field lines and hence no emf will induce. Hence emf induced is zero in this case.
Answer:
Given
Length od the solenoid
Area of the cross-section of the solenoid
Number of turns in the solenoid
Current flowing in the solenoid
The time interval for which current flows
Now.
Initial flux:
Final flux: since no current is flowing,
Now
Induced emf:
Hence 6.55V of average back emf is induced.
Answer:
Here let's take a small element dy in the loop at y distance from the wire
Area of this element dy :
The magnetic field at dy (which is y distance away from the wire)
The magnetic field associated with this element dy
Now As we know
where M is the mutual inductance
so
Hence mutual inductance between the wire and the loop is:
Answer:
Given,
Current in the straight wire
= 50A
Speed of the Loop which is moving in the right direction
Length of the square loop
distance from the wire to the left side of the square
NOW,
Induced emf in the loop :
Hence emf induced is .
Answer:
Given,
The radius of the wheel =R
The mass of the wheel = M
Line charge per unit length when the total charge is Q
Magnetic field :
Magnetic force is balanced by centrifugal force when is the speed of the wheel that is
Angular velocity of the wheel
when
It is the angular velocity of the wheel when the field is suddenly shut off.
In the Class 10th NCERT book, a brief idea of Electromagnetic Induction is given. Physics chapter 6 Class 12 give a more detailed explanation of Electromagnetic Induction and the NCERT solutions for Class 12 Physics Chapter 6 list some questions and solutions on the concepts covered in ch 6 Physics Class 12. Questions from the following topics are covered in the Electromagnetic Induction Class 12 solutions.
The experiment of Faraday and Henry, magnetic flux and Faraday's laws related to Electromagnetic Induction- This part of chapter 6 Class 12 Physics discuss the relation between magnetic field, flux and emf through experiments and equations.
Lenz's law- This law gives the polarity of induced emf.
Motional emf and its equation
Lenz's law and laws of conservation of energy-This section of ch 6 Physics Class 12 give quantitative relation of, force, power and energy.
Eddy current methods to reduce the effect of eddy current
Concept of self and mutual induction and its quantitative relation
AC generator and it's working.
NCERT solutions for class 12 physics chapter wise
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
Following are the important formulas covered in the chapter 6 physics class 12 ncert solutions Electromagnetic Induction-
Relation between flux and magnetic field:
Induced emf and magnetic field:
Motional emf: (B is perpendicular to v)
Self-induced emf:
Self-inductance:
Motional emf(ac generator):
The Class 12 Physics ch 6 NCERT solutions, is not only important for CBSE Class 12 board exams but also for many competitive exams like KVPY, NEET and JEE Main.
For exams like NEET and JEE Mains questions based on finding the direction of induced emf are expected. There are many such questions in the Electromagnetic Induction NCERT solutions.
On average 10% of questions are asked in the CBSE board exam from the chapters Electromagnetic Induction and alternating current.
Comprehensive Understanding: These solutions comprehensively cover all the topics and questions in the chapter, ensuring a deep understanding of electromagnetic induction.
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Effective Practice: Chapter 6 physics class 12 ncert solutions included in the solutions enable students to practice and assess their comprehension effectively.
Exam Readiness: These class 12 physics chapter 6 ncert solutions are an essential tool for board exam preparation and are also beneficial for competitive exams.
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These features collectively make class 12 physics electromagnetic induction ncert solutions a valuable resource, facilitating exam success and future academic pursuits.
NCERT Exemplar Class 12 Solutions
NCERT solutions subject wise:
Yes, the chapter is helpful for higher studies in the field of electronics and electrical engineering-related subjects and for scientific research.
On an average 5 mark questions are asked for CBSE board exams. Follow the NCERT syllabus and NCERT book for a good score in the board exam. Along with the NCERT solutions exercises can also practice NCERT exemplar problems and CBSE board previous year papers.
class 12 physics ch 6 ncert solutions for electromagnetic induction can provide helpful resources for students preparing for the JEE Main exam. They offer detailed explanations and solutions to the problems in the textbook, which can help students understand the concepts better and identify areas where they need more practice.
Faraday's law of electromagnetic induction states that the induced emf in a circuit is equal to the rate of change of the magnetic field times the number of turns in the circuit. Lenz's law is related to Faraday's law and states that the direction of the induced current will be such that it opposes the change that caused it. In other words, it states that the induced current will always flow in such a way as to oppose the change that caused it.
The important topics covered under electromagnetic induction ncert solutions are:
The experiment of Faraday and Henry, magnetic flux and Faraday's laws related to Electromagnetic Induction, Lenz's law and laws of conservation of energy, Eddy current methods to reduce the effect of eddy current, Concept of self and mutual induction and its quantitative relation and AC generator and it's working.
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If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
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Hello Akash,
If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.
You can get the Previous Year Questions (PYQs) on the official website of the respective board.
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Hello student,
If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:
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