NCERT Class 11 Physics Chapter 4 : Moving Charges and Magnetism
NCERT Solutions for Class 11 Physics Chapter 4 deals with Moving charges and magnetism. The interaction of moving charges with magnetic fields has been discussed. Therefore, the chapter is very essential since it covers crucial aspects about how electric currents produce magnetic fields and principles of operation of electromagnetism. The given solutions explain the core concepts associated with the Biot-Savart Law, Ampere's Law, and magnetic fields due to a current-carrying conductor. It is very important for students studying Moving Charges and Magnetism in Class 12. NCERT sets the solutions for Class 11th Physics chapter 4, which concern moving charges and the magnetism of the field. It also contains the Lorentz force, the force experienced by a moving charge in a magnetic field. Relatively very detailed derivations and examples give the students an insight into the principles of electromagnetism and how to apply them in different scenarios. These solutions are very important to master the content of Class 12 physics Chapter 4.
Download PDF For NCERT Solutions for Physics Moving Charges and Magnetism
The NCERT Class 11 Physics Chapter 4 : Moving Charges and Magnetism are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Access Answers to NCERT Class 11 Physics Chapter 4 : Moving Charges and Magnetism
Students can access the NCERT Class 11 Physics Chapter 4 : Moving Charges and Magnetism. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Physics much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Moving Charges and Magnetism
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?
Given:
Current through the wire, I = 90 A (East to West)
Distance of point P below the wire, d = 1.5 m
Direction of magnetic field,
We know that wire carries current in east to west direction. Using Right hand thumb rule, we can conclude that the direction of magnetic field is from north to south as indicated in the figure.
The magnitude of the magnetic field is 1.2 × 10-5T and its direction is from north to south.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?
Current in the wire, I = 8 A
Magnitude of the uniform magnetic field, B = 0.15 T
Angle between the wire and magnetic field, θ = 30°.
Magnetic force per unit length on the wire is given as:
f = BI sinθ
= 0.15 × 8 ×1 × sin30°
= 0.6 N m–1
Hence, the magnetic force per unit length on the wire is 0.6 N m–1.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
Given:
Number of turns, n = 100
Radius of coil, r = 8 cm
Current through the coil, I = 0.40 A
Magnitude of magnetic field at centre of coil, B = ?
⇒ |B| = 3.14 × 10-4T
∴ Magnitude of magnetic field at the centre of the coil is 3.14 × 104T.
A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
Given:
Current through the wire, I = 35 A
Distance of point P from the wire, d = 20 cm
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.
Given:
Current through the wire, I = 50 A (North to South)
Distance of point P East of the wire, d = 2.5 m
Direction of magnetic field,
The point is in a plane normal to the wire and the wire carries current in north to south. Using Right hand thumb rule we can conclude that the direction of magnetic field is vertically upwards, or out of the paper.
The magnitude of the magnetic field is 4 × 10-6T and its direction is upwards or out of paper.
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?
Length of the wire, l = 3 cm = 0.03 m
Current flowing in the wire, I = 10 A
Magnetic field, B = 0.27 T
Angle between the current and magnetic field, θ = 90°
Magnetic force exerted on the wire is given as:
F = BIlsinθ
= 0.27 × 10 × 0.03 sin90°
= 8.1 × 10–2 N
Hence, the magnetic force on the wire is 8.1 × 10–2 N. The direction of the force can be obtained from Fleming’s left hand rule.
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.
Given:
Current in wire A, IA = 8.0 A
Current in wire B, IB = 5.0 A
Distance between the conductors A and B, d = 4 cm
Length of conductor on which we have to calculate force, L = 10cm
So, the force on the 10 cm section on wire A is 2 × 10-5N. Since the current is flowing in the same direction the force will be attractive in nature.
Note: The force will be same on both the wires, we can use Newton’s third law of motion to such conclusion.
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 carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Given:
Length of solenoid, L = 80cm
Number of turns = number of layers × number of turns per layer
Number of turns, n = 5 × 400 = 2000
Radius of solenoid, r = Diameter/2 = 0.9 cm
Current through the solenoid = 8.0A
Hence the magnetic field strength at the centre of the solenoid is 2.512 × 10-2T.
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 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Given:
Length of side of square, L = 10 cm
Number of turns, n = 20
Current through the square coil, I = 12 A
Angle between the normal to the coil and uniform magnetic field, θ = 30°
Magnitude of magnetic field, B = 0.80 T
Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, 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 the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.
Given:
For moving coil meter M1
Resistance of wire, R1 = 10Ω
Number of turns, N1 = 30
Area of cross-section, A1 = 3.6 × 10-3 m2
Magnetic field strength, B1 = 0.25 T
For moving coil meter M2
Resistance of wire, R2 = 14Ω
Number of turns, N2 = 42
Area of cross-section, A2 = 1.8 × 10-3 m2
Magnetic field strength, B2 = 0.50 T\
Spring constant, K1 = K2 = K
Current sensitivity is given by,
Hence, the ratio of current sensitivities is 1.4.
Hence, the ratio of voltage sensitivity of M1 and M2 is 1.
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)
Given:
Magnetic field strength, B = 6.5 G = 6.5 × 10-4T
Initial velocity of electron = 4.8 × 106 ms-1
Angle between the initial velocity of electron and magnetic field, θ = 900
⇒ Fe = 1.6 × 10-19 C × 4.8 × 106 ms-1 × 6.5 × 10-4T × sin 90
⇒ Fe = 4.99 × 10-16N
This force serves as the centripetal force, which explains the circular trajectory of the electron.
Centripetal force Fc = mv2/r …(2)
By equating equation (1) and equation (2) we get,
In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Given:
Magnetic field strength, B = 6.5 G = 6.5 × 10-4T
Initial velocity of electron = 4.8 × 106 ms-1
Angle between the initial velocity of electron and magnetic field, θ = 900
We can relate the velocity of the electron to its angular frequency by the relation,
V = rω …(1)
Where,
V = velocity of electron
r = radius of path
ω = angular frequency
(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 60º 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.)
Given:
Number of turns in the coil, n = 30
Radius of coil, r = 8 cm
Current through the coil, I = 6.0 A
Strength of magnetic field = 1.0 T
Angle between the direction of field and normal to coil, θ = 60°
We can understand that the counter torque required to prevent the coil from rotating is equal to the torque being applied by the magnetic field.
Torque on the coil due to magnetic field is given by,
T = n × B × I × A × sinθ …(1)
Where,
n = number of turns
B = Strength of magnetic field
I = Current through the coil
A = Area of cross-section of coil
A = πr2 = 3.14 × (0.08 × 0.08) = 0.0201m2 …(2)
θ = Angle between normal to cross-section of coil and magnetic field
Now, by putting the values in equation (1) we get,
⇒ T = 30 × 6.0T × 1A × 0.0201m2 × sin60°
T = 3.133 Nm
Hence, the counter torque required to prevent the coil from rotating is 3.133 Nm.
b) From equation (1) we can understand that torques depend on the total area of cross-section and has no relation with the geometry of cross-section. Hence, the answer will remain unaltered if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area.
Moving Charges and Magnetism - ADDITIONAL EXERCISES
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.
Here we have to find the total magnetic field produced by the system so we will first find the magnetic field due to each coil with direction and then add them in accordance with vector addition. Using the Right-hand thumb rule we can predict the direction of the induced magnetic field in both the coils.
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