NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism: Students who are preparing for class 12 board exam must go through Moving Charges and Magnetism NCERT solutions in this article. By going through Moving Charges and Magnetism NCERT solutions, students will be able to understand the core concepts of this topic. Moving Charges and Magnetism include important topics like Magnetic Field and direction of a Circular Coil, Magnetic Force, etc. Moving Charges and Magnetism Class 12 notes will also help the students to prepare for competitive exams like JEE and NEET. The subject matter experts have created Moving Charges and Magnetism NCERT solutions to help the students to prepare well for the exam and secure a good score. 

Check Out: CBSE Class 12th Books

Moving Charges and Magnetism NCERT Solutions

Students can check the questions and answers of Moving Charges and Magnetism NCERT solutions in below

Question 1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.4 A. What is the magnitude of the magnetic field 6 at the centre of the coil?
Solution:
The magnetic field at the centre of a circular coil having 100 turns.

Read More: NCERT Solutions For Class 12 Physics Chapter 1 Electric Charges and Fields

Question 2. A long straight wire carries a current of 35 A. What is the magnitude of the field 6 at a point 20 cm from the wire?
Solution:
Magnetic field due to a long straight wire

Read More: NCERT Solutions For Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Question 3. A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Solution:
Let us first decide the standard directions on the plane of paper.

Magnitude of magnetic field

Read More: NCERT Solutions For Class 12 Physics Chapter 3 Current Electricity

Question 4. A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Solution:
The standard directions on the plane of paper can be different according to requirements.

Direction of magnetic field can be observed by right hand palm rule and it is southward.

Question 5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?
Solution:

Let l be length of wire, carrying a current of 8 A at an angle 30° with the magnetic field. Force on the wire, F = |B| sinθ Force per unit length F/l = |B| sinθ F/l = 8 × 0.15 × sin 30° = 0.6 N m-1.

Question 6. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Solution:
The magnetic field inside the solenoid is along its axis. the current in the wire flows perpendicular to the axis. F = |B| sin 90° or F = 10 × 0.27 × 3 × 10-2 x 1 = 0.081 N

Question 7. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Solution:
Force of attraction per unit length on two parallel wires carrying current in same direction.

Question 8. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current the carried is 8.0 A, estimate the magnitude of 8 inside the solenoid near its centre.
Solution:
Total number of turns in 80 cm length of solenoid can be calculated

Question 9. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Solution:

Torque experienced by the coil carrying current in the given magnetic field.

Question 10. Two moving coil meters, M1 and M2 have following particulars: R1 = 10Ω, N1 = 30, A1 = 3.6 × 10-3 m2, B1 = 0.25 T R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10-3 m2, B2 = 0.50 T. (The spring constants are identical for two meters). Determine ratio of (i) current sensitivity (ii) voltage sensitivity of M2 and A4,
Solution:
Current sensitivity of a moving coil galvanometer is defined as

(ii) Ratio of voltage sensitivity

Question 11. In a chamber, a uniform magnetic field of 6.5 G (1 G = 10-4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e= 1.6 × 10-19 C,me = 9.1 × 10-31 kg)
Solution:
The magnetic force ƒ = quB act normal to the direction of motion, thus provide the necessary centripetal force to follow the circular path.

Question 12. In question 11, obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Solution:
Time period can be calculated.

The frequency of revolution of electron is independent of speed of electron.

Question 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 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? [All other particulars are also unaltered].
Solution:
(a) The given coil is circular and is suspended such that field lines makes angle 60° with normal of the coil.A similar torque is required to prevent the coil from turning.

A similar torque is required to prevent the coil from turning.
(b) As long as the area of the planar coil remains same, the torque on the coil is also same, irrespective of the shape.

Question 14. Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
Solution:
The concentric coils are in the plane north to south. Let us decide the directions with conveniences in the plane of paper.

Magnetic field at centre due to coil X

So, a magnetic field nearly 1.6 × 10-3 T will appear at centre of the circular coil directed towards west.

Question 15. A magnetic field of 100 G (1G = 10-4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 103-3 m2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m”1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Solution:
The magnetic field required is 100 G in a length of 10 cm. So, the solenoid should have a length larger than 10 cm. The maximum current capacity of wire is 15 A, so less than 15 A should flow in wire. Now using

So, one combination for the desired 100 G magnetic field in a length 10 cm can be a total length of solenoid of 50 cm and current in the solenoid of 10 A and total turns nearly 400.

Check out: Class 12th Sample Papers

Question 16. Fora 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

(a) Show that this reduces to the familiar result for field at the centre of the coil
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
Solution:
On the axis of the circular coil of radius R and N turns carrying a current I at a distance x from centre the magnetic field is

Which is the same result as is obtained by integration at centre of the coil.
(b) The side of the loop where current flows clockwise becomes south pole and where the current appear anticlockwise becomes north pole. Direction of magnetic field is from south pole to north pole.

Now, two parallel coils each of radius R with N turns. The magnetic field of both the coils add.

Question 17. A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field
(a) outside the toroid,
(b) inside the core of the toroid, and
(c) in the empty space surrounded by the toroid.
Solution:

Total turns in the toroid are given 3500.
(a) outside the toroid the magnetic field is zero.
(b) Inside the core of toroid, the magnetic field will be

(c) Magnetic field in the empty space surrounded by the toroid is also zero.

Question 18. Answer the following questions:
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.
Solution:
(a) If a charged particle move parallel or anti parallel to the magnetic field, no magnetic force will act on it and it move undeflected. So, in the given condition either the charged particle enters east to west or west to east.
(b) A magnetic force can only change the direction of charged particle but never changes magnitude of speed as force act normal to direction of speed. So, charged particle may follow a complicated trajectory, but its speed remains the same.
(c)

If,we want the electron to move undeflected in the presence of electric and magnetic fields, then the electric force should be balanced by magnetic force.

So, the magnetic field should act in downward direction.

Question 19. An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
Solution:
A electron which is accelerated by a potential difference of 2.0 kV will have a kinetic energy gained 2000 eV.

(a) When the electron enters in the uniform r magnetic field which is normal to the velocity of electron the electron follows a circular path r of radius.

(b) When the magnetic field makes an angle 30° with the initial velocity, the trajectory of the electron becomes helical. radius of the helical path is

Question 20. A magnetic field set up using Helmholtz coils (described in question 16 above) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10-5 V m-1, make a simple guess as to what the beam contains. Why is the answer not unique?
Solution:
Narrow beam of charged particles remains undeflected and is perpendicular to both electric field and magnetic fields which are mutually perpendicular. So, the electric force is balanced by magnetic force. qE = quB Speed of charged particles

Because the beam is accelerated through 15 kV, if charge is q, then kinetic energy gained by charged particles

Here, we can only obtain charge to mass ratio and same ratio can be in Deuterium ions, He++, Li+++, so the beam can contain any of these charged particles.

Check Out: Class 12th Question Banks

Moving Charges and Magnetism Class 12 Notes Summary

The chapter begins by explaining the relationship between electricity and magnetism. When an electric current flows through a wire, it generates a magnetic field around it. This is demonstrated using the right-hand rule: if you point your thumb in the direction of the current, your curled fingers show the direction of the magnetic field.

Next, the chapter covers Ampère’s Circuital Law, which helps us understand how to find the magnetic field produced by a current-carrying conductor. It states that the magnetic field around a conductor is proportional to the current and inversely proportional to the distance from the conductor.

The chapter then introduces the concept of the Magnetic Force on a Current-Carrying Conductor. When a current-carrying wire is placed in a magnetic field, it experiences a force. This force can be calculated using the formula F=I⋅L⋅B⋅sin⁡θ, where I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the current and the magnetic field.

The chapter also discusses the moving charges in a magnetic field. When a charged particle moves through a magnetic field, it experiences a force called the Lorentz force. This force is perpendicular to both the velocity of the particle and the magnetic field.

Finally, the chapter covers applications like electric motors and generators, which rely on these principles to convert electrical energy into mechanical energy and vice versa. Understanding these concepts helps in grasping how many electrical devices work.

Moving Charges and Magnetism Class 12 Notes FAQs

Q1. What is the name of Chapter 4 of Class 12 physics?

Ans. The name of Chapter 4 of Class 12 physics is Moving Charges and Magnetism.

Q2. What are the important topics from moving charges and magnetism class 12?

Ans. The important topics from moving charges and magnetism are Magnetic Field Caused by a Current Element, Biot-Savart Law. Magnetic Field and Magnetic Force. Ampere's Circuital Law.

Q3. What are the topics in Moving Charges and Magnetism?

Ans. The topics in Moving Charges and Magnetism includes Magnetic Force, Motion in a Magnetic Field, Motion in Combined Electric and Magnetic Fields, Magnetic Field due to a Current Element, Biot-Savart Law, Magnetic Field on the Axis of a Circular Current Loop, Ampere’s Circuital Law, The Solenoid and the Toroid, Force between Two Parallel Currents, the Ampere, Torque on Current Loop, Magnetic Dipole, Torque on Current Loop, Magnetic Dipole. 

Q4. Which is the easiest chapter in physics class 12?

Ans. The easiest chapters in Physics class 12 are Atoms, Nucleus, Communication (if it still remains in syllabus), Electromagnetic Waves (last chapter of book one., Dual matter and radiation.