NCERT Solutions for Class 12 Physics Chapter 4
NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism chapter end Exercises Questions Answers and Additional Exercises Solutions in PDF format updated for academic session 2020-21.UP Board Solutions, NCERT Solutions and CBSE Offline Apps are updated according to new CBSE Books for 2020-21.
NCERT Solutions for Class 12 Physics Chapter 4
|Chapter 4:||Moving Charges and Magnetism|
Chapter 4 Moving Charges and Magnetism Solutions
- Study Online: Exercises Question Answers
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- Practice Questions for Competitive Exams
- NCERT Book Chapter 4
- NCERT Book Answers
- Revision Book Chapter 4
- Revision Book Answers
- Revision Notes 1
- Revision Notes 2
- Visit to 12th Physics Main Page
Class 12 Physics Chapter 4 Solutions in English
NCERT Solutions for Class 12 Physics Chapter 4 in PDF form to free download for academic session 2020-21. Ask your doubts related to NIOS or CBSE Board through Discussion Forum and response to the questions asked by others. Download NCERT Books 2020-21 based on latest CBSE Syllabus for UP Board, MP Board and other boards who are following CBSE Curriculum.
Important Questions for practice
1. An electric current flows in a horizontal wire from East to West. What will be the direction of magnetic field due to current at a point (i) North of wire; (ii) above the wire.
2. If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis, which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge).
3. An electron beam projected along + x-axis, experiences a force due to a magnetic field along the + y-axis. What is the direction of the magnetic field?
4. In a circuit instantaneously voltage and current are V = 150 sin 314t volt and i = 12 cos 314 t ampere respectively. Is the nature of circuit is capacitive or inductive?
5. A wire moves with some speed perpendicular to a magnetic field. Why is emf induced across the rod?
Questions from Board Papers
1. Define RMS Value of Current. [Answer: RMS Value of ac is defined as that value of direct current which produces the same heating effect in a given resistor as is produced by the given alternating current when passed for the same time.]
2. A galvanometer of resistance 120 ohms gives full scale deflection for a current of 5mA. How can it be converted into an ammeter of range 0 to 5A? Also determine the net resistance of the ammeter.
3. A proton and an alpha particle of the same enter, in turn, a region of uniform magnetic field acting perpendicular to their direction of motion. Deduce the ratio of the radii of the circular paths described by the proton and alpha particle.
4. A magnetic dipole of magnetic moment M is kept in a magnetic field B. What is the minimum and maximum potential energy? Also give the most stable position and most unstable position of magnetic dipole.
5. A circular coil of n turns and radius R carries a current I. It is unwound and rewound to make another square coil of side ‘a’ keeping number of turns and current same. Calculate the ratio of magnetic moment of the new coil and the original coil.
Important Questions on 12th Physics Chapter 4
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
Hence, the magnetic force per unit length on the wire is 0.6 N/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.
Current flowing in the coil, I = 12 A
Number of turns on the coil, n = 20
Angle made by the plane of the coil with magnetic field, θ = 30°
Strength of magnetic field, B = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
τ = n BIA sinθ
Where, A = Area of the square coil
= l × l = 0.1 × 0.1
= 0.01 m2
τ = 20 × 0.8 × 12 × 0.01 × sin30°
= 0.96 N m
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.
Radius of the coil, r = 8.0 cm = 0.08 m
Area of the coil πr^2=π(0.08)^2=0.0201 m^2
Current flowing in the coil, I = 6.0 A
Magnetic field strength, B = 1 T
Angle between the field lines and normal with the coil surface, θ = 60°
The coil experiences a torque in the magnetic field.
Hence, it turns.
The counter torque applied to prevent the coil from turning is given by the relation,
τ = n IBA sin θ
= 30 × 6 × 1 × 0.0201 × sin60°
= 3.133 N m