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The NCERT solutions provide step-by-step explanations for all the problems in Chapter 4, helping you understand the application of laws. With the help of the NCERT solution for Class 12 Physics, students may ensure that they have a solid understanding of the concepts presented in each chapter and prepare for tests such as JEE Main, NEET and CBSE Class 12.
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The NCERT solutions include solved exercise questions, additional questions and HOTS, or higher order thinking skills questions, to create higher order problem solving skills, important Topics and formulas of magnetic fields, Biot-Savart Law, Ampere Circuital Law, force on current carrying conductor, and magnetic field due to a current loop and approach to solve questions.
The important topics in Class 12 Physics Chapter 4, Moving Charges and Magnetism, are the Lorentz Force and the Motion of a Charged Particle, Biot-Savart Law and its Applications, Ampere's Circuital Law and its Applications, Force Between Two Parallel Current-Carrying Wires, and Moving Coil Galvanometer
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You will find complete step-by-step solutions of NCERT Class 12 Physics Chapter 4 – Moving Charges and Magnetism. Get the free PDF from us to enhance your knowledge base and improve your test performance.
Answer :
The magnitude of the magnetic field at the centre of a circular coil of radius r carrying current I is given by,
For 100 turns, the magnitude of the magnetic field will be,
(Given, current=0.4A, radius = 0.08m, permeability of free space = 4
Answer:
The magnitude of the magnetic field at a distance r from a long straight wire carrying current I is given by,
In this case
(Given, current=0.4A, radius = 0.08m, permeability of free space = 4
Answer:
The magnitude of the magnetic field at a distance r from a long straight wire carrying current I is given by,
In this case
(Given, current=0.4A, radius = 0.08m, permeability of free space = 4
The current is going from the North to South direction in the horizontal plane and the point lies to the East of the wire. Applying Maxwell's right-hand thumb rule we can see that the direction of the magnetic field will be vertically upwards.
Answer:
The magnitude of the magnetic field at a distance r from a long straight wire carrying current I is given by,
In this case
(Given, current=0.4A, radius = 0.08m, permeability of free space = 4
The current in the overhead power line is going from the East to West direction and the point lies below the power line. Applying Maxwell's right-hand thumb rule we can see that the direction of the magnetic field will be towards the South.
Answer:
The magnetic force on an infinitesimal current-carrying conductor in a magnetic field is given by
For a straight wire of length l in a uniform magnetic field, the Force equals to
In the given case the magnitude of force per unit length is equal to
(Given, I=8A, B=0.15 T,
|F| = 0.15
Answer:
For a straight wire of length l in a uniform magnetic field, the Force equals to
In the given case the magnitude of the force is equal to
(Given, I=10A, B=0.27 T,
|F| = 0.27
The direction of this force depends on the orientation of the coil and the current-carrying wire and can be known using the Flemings Left-hand rule.
Answer:
The magnitude of magnetic field at a distance r from a long straight wire carrying current I is given by,
In this case the magnetic field at a distance of 4.0 cm from wire B will be
(Given, I=5 A, r=4.0 cm)
The force on a straight wire of length l carrying current I in a uniform magnetic field B is given by
The force on a 10 cm section of wire A will be
(Given, B=2.5 T, I=8 A, l = 10 cm,
Answer:
The magnitude of the magnetic field at the centre of a solenoid of length
In the given question N= number of layers of winding
N=5
I=8.0 A
Answer:
The magnitude of torque experienced by a current-carrying coil in a magnetic field is given by
where n = number of turns, I is the current in the coil, A is the area of the coil and
In the given question n = 20, B=0.8 T, A=0.1
The coil, therefore, experiences a torque of magnitude 0.96 Nm.
4.10 (a) Two moving coil meters,
(The spring constants are identical for the two meters).
Determine the ratio of current sensitivity of
Answer:
The torque experienced by the moving coil M1 for a current I passing through it will be equal to
The coil will experience a restoring torque proportional to the twist
The current sensitivity is therefore
Similarly, for the coil M2, current sensitivity is
Their ratio of current sensitivity of coil M2 to that of coil M1 is, therefore,
4.10(b) Two moving coil meters,
(The spring constants are identical for the two meters).
Determine the ratio of voltage sensitivity of
Answer:
The torque experienced by the moving coil M1 for a current I passing through it will be equal to
The coil will experience a restoring torque proportional to the twist
we know V=IR
Therefore,
Voltage sensitivity of coil M1 =
Similarly for coil M2 Voltage sensitivity =
Their ratio of voltage sensitivity of coil M2 to that of coil M1 is:
Answer:
The magnetic force on a moving charged particle in a magnetic field is given by
Since the velocity of the shot electron is perpendicular to the magnetic field, there is no component of velocity along the magnetic field and therefore the only force on the electron will be due to the magnetic field and will be acting as a centripetal force causing the electron to move in a circular path. (if the initial velocity of the electron had a component along the direction of the magnetic field it would have moved in a helical path)
Magnetic field(B)=
Speed of electron(v)=
Charge of electron=
Mass of electron=
The angle between the direction of velocity and the magnetic field = 90o
Since the force due to the magnetic field is the only force acting on the particle,
Answer:
In exercise 4.11 we saw
Time taken in covering the circular path once(time period (T))=
Frequency,
From the above equation, we can see that this frequency is independent of the speed of the electron.
4.13 (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of
Answer:
Number of turns in the coil(n)=30
The radius of the circular coil(r)=8.0 cm
Current flowing through the coil=6.0 A
Strength of magnetic field=1.0 T
The angle between the field lines and the normal of the coil=60o
The magnitude of the counter-torque that must be applied to prevent the coil from turning would be equal to the magnitude of the torque acting on the coil due to the magnetic field.
A torque of magnitude 3.13 Nm must be applied to prevent the coil from turning.
Answer:
From the relation
Answer:
Using the right-hand thumb rule we can see that the direction of the magnetic field due to coil X will be towards the east direction and that due to coil Y will be in the West direction.
We know the magnetic field at the centre of a circular loop of radius r carrying current I is given by
The net magnetic field at the centre of the coils,
Bnet =By - Bx =1.57
The direction of the magnetic field at the centre of the coils is towards the west direction.
Answer:
Strength of the magnetic field required is
Therefore keeping the number of turns per unit length and the value of current within the prescribed limits such that their product is approximately 8000 we can produce the required magnetic field.
e.g. n=800 and I=10 A.
Show that this reduces to the familiar result for the field at the centre of the coil.
Answer:
For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,
For finding the field at the centre of coil we put x=0 and get the familiar result
Answer:
Let a point P be at a distance of l from the midpoint of the centres of the coils.
The distance of this point from the centre of one coil would be
The magnetic field at P due to one of the coils would be
The magnetic field at P due to the other coil would be
Since the direction of current in both the coils is same the magnetic fields B1 and B2 due to them at point P would be in the same direction
Since
Since the above value is independent of
Answer:
Outside the toroid, the magnetic field will be zero.
Answer:
The magnetic field inside the core of a toroid is given by
Total number of turns(N)=3500
Current flowing in toroid =11 A
Length of the toroid,
Answer:
The magnetic field in the empty space surrounded by the toroid is zero.
Answer:
The charged particle is not deflected by the magnetic field even while having a non zero velocity, therefore, its initial velocity must be either parallel or anti-parallel to the magnetic field i.e. It's velocity is either towards the east or the west direction.
Answer:
Yes, its final speed will be equal to the initial speed if it has not undergone any collision as the work done by the magnetic field on a charged particle is always zero because it acts perpendicular to the velocity of the particle.
Answer:
The electron would experience an electrostatic force towards the north direction, therefore, to nullify its force due to the magnetic field must be acting on the electron towards the south direction. By using Fleming's left-hand rule we can see that the force will be in the north direction if the magnetic field is in the vertically downward direction.
Explanation:
The electron is moving towards the east and has a negative charge therefore
Answer:
The electron has been accelerated through a potential difference of 2.0 kV .
Therefore K.E of electron
Since the electron initially has velocity perpendicular to the magnetic field it will move in a circular path.
The magnetic field acts as a centripetal force. Therefore,
Answer:
The electron has been accelerated through a potential difference of 2.0 kV .
Therefore K.E of electron
The component of velocity perpendicular to the magnetic field is
The electron will move in a helical path of radius r given by the relation,
The component of velocity along the magnetic field is
The electron will move in a helical path of pitch p given by the relation,
The electron will, therefore, move in a helical path of radius 5 mm and pitch 5.45 mm.
Answer:
Let the beam consist of particles having charge q and mass m.
After being accelerated through a potential difference
Using the value of
Answer:
In order for the tension in the wires to be zero the force due to the magnetic field must be equal to the gravitational force on the rod.
mass of rod=0.06 g
length of rod=0.45m
the current flowing through the rod=5 A
A magnetic field of strength 0.261 T should be set up normal to the conductor in order that the tension in the wires is zero
Answer:
If the direction of the current is reversed the magnetic force would act in the same direction as that of gravity.
Total tension in wires(T)=Gravitational force on rod + Magnetic force on rod
The total tension in the wires will be 1.176 N.
Answer:
Since the distance between the wires is much smaller than the length of the wires we can calculate the Force per unit length on the wires using the following relation.
Current in both wires=300 A
Distance between the wires=1.5 cm
Permeability of free space=4
F=1.2 Nm-1
Answer:
The length of wire inside the magnetic field is equal to the diameter of the cylindrical region=20.0 cm=0.2 m.
Magnetic field strenth=1.5 T.
Current flowing through the wire=7.0 A
The angle between the direction of the current and magnetic field=90o
Force on a wire in a magnetic field is calculated by relation,
F=2.1 N
This force due to the magnetic field inside the cylindrical region acts on the wire in the vertically downward direction.
Answer:
Magnetic field strenth=1.5 T.
Current flowing through the wire=7.0 A
The angle between the direction of the current and magnetic field=45o
The radius of the cylindrical region=10.0 cm
The length of wire inside the magnetic field,
Force on a wire in a magnetic field is calculated by relation,
F=2.1 N
This force due to the magnetic field inside the cylindrical region acts on the wire in the vertically downward direction.
This force will be independent of the angle between the wire and the magnetic field as we can see in the above case.
Note: There is one case in which the force will be zero and that will happen when the wire is kept along the axis of the cylindrical region.
Answer:
The wire is lowered by a distance d=6cm.
In this case, the length of the wire inside the cylindrical region decreases.
Let this length be
This force acts in the vertically downward direction.
Answer:
The magnetic field is
Current in the loop=12 A
Area of the loop = length
A=0.1
A=0.005 m2
Torque is given by,
The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-y direction. The force on the loop is zero.
Answer:
The magnetic field is
Current in the loop=12 A
Area of the loop = length
A=0.1
A=0.005 m2
The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-y direction. The force on the loop is zero.
Answer:
The magnetic field is
Current in the loop=12 A
Area of the loop = length
A=0.1
A=0.005 m2
Torque is given by,
The torque on the loop has a magnitude of 0.018 Nm and acts along the negative-x-direction. The force on the loop is zero.
Answer:
The magnetic field is
Current in the loop=12 A
Area of the loop = length
A=0.1
A=0.005 m2
Torque is given by,
The torque on the loop has a magnitude of 0.018 Nm and at an angle of 240o from the positive x-direction. The force on the loop is zero.
Answer:
The magnetic field is
Current in the loop=12 A
Area of the loop = length
A=0.1
A=0.005 m2
Since the area vector is along the direction of the magnetic field the torque on the loop is zero. The force on the loop is zero.
Answer:
The magnetic field is
Current in the loop=12 A
Area of the loop = length
A=0.1
A=0.005 m2
Since the area vector is in the opposite direction of the magnetic field the torque on the loop is zero. The force on the loop is zero.
The force on the loop in all the above cases is zero as the magnetic field is uniform
Answer:
As we know the torque on a current-carrying loop in a magnetic field is given by the following relation
It is clear that the torque, in this case, will be 0 as the area vector is along the magnetic field only.
Answer:
The total force on the coil will be zero as the magnetic field is uniform.
Answer:
The average force on each electron in the coil due to the magnetic field will be eV d B where V d is the drift velocity of the electrons.
The current is given by
where n is the free electron density and A is the cross-sectional area.
The average force on each electron is
Answer:
The magnetic field inside the solenoid is given by
n is number of turns per unit length
n=1500 m-1
Current in the wire Iw = 6 A
Mass of the wire m = 2.5 g
Length of the wire l = 2 cm
The windings of the solenoid would support the weight of the wire when the force due to the magnetic field inside the solenoid balances weight of the wire
Therefore a current of 108.37 A in the solenoid would support the wire.
Answer:
The galvanometer can be converted into a voltmeter by connecting an appropriate resistor of resistance R in series with it.
At the full-scale deflection current(I) of 3 mA the voltmeter must measure a Voltage of 18 V.
The resistance of the galvanometer coil G =
The galvanometer can be converted into a voltmeter by connecting a resistor of resistance
Answer:
The galvanometer can be converted into an ammeter by connecting an appropriate resistor of resistance R in series with it.
At the full-scale deflection current(I) of 4 mA, the ammeter must measure a current of 6 A.
The resistance of the galvanometer coil is G =
Since the resistor and galvanometer coil are connected in parallel the potential difference is the same across them.
The galvanometer can be converted into an ammeter by connecting a resistor of resistance
Understanding class 12 NCERT solutions on moving charges and magnetism is similar to constructing a tall building's base. It is essential for passing difficult entrance examinations like JEE and NEET (like the support system that holds up the building) and for performing well on your regular school exams (like the bricks at the bottom).
Question 1: A charged particle is thrown into space with a uniform magnetic field and uniform electric field.
Choose the correct possibility of the path of the charged particle.
1) The path of a charged particle may be a straight line, cycloidal or helical.
2) The path of the charged particle may be elliptical.
3) If electric field is removed path of the charge particle must be circular
4) None of the above
Answer:
In a pure uniform electric field, any charged particle will either move in a straight line or a parabolic path.
In a pure uniform magnetic field, any charged particle will either move on a circular path or a helical path.
In the presence of an electric field and magnetic field, the charged particle may move on a straight line if magnetic force balances the electric force
This situation comes in the case of a velocity selector.
If the electric field and magnetic field are parallel then the charge particle will follow a helical path.
If the electric field and magnetic field are perpendicular to each other then the charged particle may follow the cycloidal path.
Hence, the answer is the option (1).
Question 2: A particle of charge '
1) The coordinates of the charge particle at
2) The coordinates of the charge particle at
3) The coordinates of the charge particle at
4) All are incorrect
Answer:
The particle will follow the helical path, which has a circular projection in the Y – Z plane and a straight-line motion along the x-axis.
In the Y-Z plane, it will perform circular motion with
and centre of circle is at
At general time its
Its X coordinate increases linearly with speed v.
(a) at
(b) at
(c) at
Hence, the answer is the option(1) and option(2).
Question 3: Choose the correct statement about magnetic field.
1) Any charged particle in the magnetic field must experience a force due to it.
2) Any moving charge particle in the magnetic field must experience force due to it.
3) The work done by magnetic force on a moving charge particle is always zero.
4) The work done by magnetic force on a current carrying wire maybe non-zero.
Answer:
The magnetic field can apply force on a moving charge particle if it's velocity is not parallel to magnetic field.
Result can be easily viewed on basis of Lorentz force.
It says the force is obtained by cross product of velocity of charge particle and magnetic field.
Magnetic force never does any work, neither on charge nor on current carrying wire.
Hence, the answer is the option (3).
Question 4: Two proton beams are moving parallel to each other with same velocity v along their beam direction at some separation. Find out the correct option(s):
1) Ratio of electrostatic repulsive and magnetic attraction force is
2) Ratio of electrostatic repulsive and magnetic attraction force is
3) Electrostatic force per unit length of any beam is
Magnetic force per unit length of any beam is
4) Electrostatic force per unit length of any beam is
Magnetic force per unit length of any beam is
Answer:
Let there be n protons in unit length
So charge / length,
Effective current in proton beam
Now electrostatic force per unit length of any beam
Magnetic force per unit length of any beam
Hence, the correct answer is the option(1) and (3).
Question 5: A proton, a deuterium nucleus, and an alpha particle are accelerated through the same potential difference. Now, these particles are sent in a uniform magnetic field perpendicularly. Choose the correct ratio(s):
1) Their radii of circles
2) Their angular momentum
3) Their frequencies of periodic motion
4) All the above are correct
Answer:
If charge '
In a magnetic field,
Proton, deuterium, and a particle have charge and mass as
(a) So, their ratio of radii :
(b) Their ratio of angular momentum :
(c) Their ratio of frequencies:
Hence, the answer is the option (4).
Chapter 4 Moving charges and magnetism NCERT Class 12 physics addresses the nature of magnetic fields generated by moving electric charges, the properties of such fields, how they interact with other moving charges, and with currents. It prepares the ground to study the magnetic effects of current, which play an essential role both in electromagnetism and in modern technology.
4.1 Introduction
4.2 Magnetic Force
4.2.1 Sources And Fields
4.2.2 Magnetic Field, Lorentz Force
4.2.3 Magnetic Force On A Current-Carrying Conductor
4.3 Motion In A Magnetic Field
4.4 Magnetic Field Due To A Current Element, Biot-Savart Law
4.5 Magnetic Field On The Axis Of A Circular Current Loop
4.6 Ampere’S Circuital Law
4.7 The Solenoid
4.8 Force Between Two Parallel Currents, The Ampere
4.9 Torque On Current Loop, Magnetic Dipole
4.9.1 Torque On A Rectangular Current Loop In A Uniform Magnetic Field
4.9.2 Circular Current Loop As A Magnetic Dipole
4.10 The Moving Coil Galvanometer
The important formulas from Class 12 Physics Chapter 4 – Moving Charges and Magnetism are essential for solving exercise questions accurately. These formulas help in understanding magnetic force, field calculations, and motion of charged particles in magnetic fields.
Where:
The force is perpendicular to both the velocity and magnetic field.
Where:
Where:
Where:
Where:
Where:
Where:
Moving Charges and Magnetism is basically about how moving electric charges can generate a magnetic field. It also explains the magnetic force and nature of a magnetic field. There are two important laws discussed in the chapter as well: the Ampere Law and the Biot-Savart Law. The laws describe the behaviour of moving charges in terms of current-carrying conductors as the cause of creating magnetic fields. An example is when you switch on an electric fan and the wires give the current that generates a magnetic field, consequently causing the motor of the fan to spin. Know what the approach should be to cover this chapter.
Learn about the Concept First
Make Neat Diagram
Apply Right-Hand Rules
Determine the required Formula
While NCERT provides a strong foundation, students preparing for JEE and NEET need to go beyond it for advanced concepts and practice. The comparison table below highlights the additional study requirements for each exam beyond the NCERT syllabus.
As the CBSE board exam is concerned, the solutions of NCERT Class 12 Physics chapter 4 Moving Charges and Magnetism is important. In CBSE board exams, 12 % of questions are asked from chapters 4 and 5. Same questions discussed in the chapter 4 Physics Class 12 NCERT solutions can be expected in the board exams.
Here are the exercise-wise solutions of the NCERT Class 12 physics book:
Frequently Asked Questions (FAQs)
The chapter Moving Charges and Magnetism have 8 to 10 percentage weightage. The questions asked from the chapter can be of a numerical, derivation or theory questions. CBSE board follows NCERT Syllabus. To practice problems refer to NCERT text book, NCERT syllabus and previous year board papers of Class 12 Physics.
No, you can not skip it. Since from NCERT Class 12 Physics chapter 4 you can expect 2 questions for NEET exam.
No, you have to practise more questions for doing well in JEE main exams. The questions of Moving Charges and Magnetism need a good base of vectors and a thorough understanding of concepts
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Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
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