NCERT Solutions For Class 12 Physics Chapter 4: Moving Charges and Magnetism

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NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism are provided in this article. Moving charges generate an electric field. The rate of flow of electric charge is known as current. Magnetism is caused due to the current. Magnetic fields exert forces on the magnets and the moving charges.

Class 12 Physics Chapter 4 along with Chapter 5 Magnetism and Matter belongs to Unit 3 which has a weightage of 17 marks with Unit 4 Electromagnetic Induction and Alternating Currents. Along with the elementary concepts, mathematical treatment of the magnetic field produced due to a current element (Biot Savart Law), ampere’s circuital law and the concept of solenoid and toroids are covered in the Class 12 NCERT Solutions for NCERT Solutions for Class 12 Physics Chapter 4.

Download PDF: NCERT Solutions for Class 12 Physics Chapter 4

NCERT Solutions for Class 12 Physics Chapter 4

The NCERT solutions for class 12 physics chapter 4: Moving Charges and Magnetism are provided below.

NCERT SOLUTIONS

Class 12 Physics Chapter 4 – Important Concepts

The region in space where a Magnet has its magnetic effect is known as the Magnetic field of the Magnet.

F = q [E(r) + v × B(r)] = E_Electric + F_magnetic

Magnetism is a property displayed by Magnets and produced by the moving charges. This results in the objects being attracted or pushed away.

The relation between a moving charge and magnetism is that Magnetism is caused due to the movement of charges.

Lorentz Force is the total force on a given charge c, moving with a velocity v, in the presence of electric field E and magnetic field B. (This force acts normal to v and the work done by it is zero)

F = q(v x B + E)

Magnetic Force on a Current-Carrying Conductor: Due to the motion of charges in a conductor, each charge experiences a force. When current is passed through a magnetic field, the magnetic field exerts a force on the wire in a perpendicular direction to the current and the magnetic field as well.

$F= I (I \times B ) or |F | = I |I||B| sin \theta $

Also Read:

Chapter 4 Related Articles
Lorentz Force	Unit of Magnetic Field	Solenoid and Toroids
Ampere’s Circuital Law	Biot Savart Law	Electromagnetic Field
Electrical Formula	Magnetic Field	The Moving Coil Galvanometer
Magnetic Force and Magnetic Field	Magnetic Field on the Axis of a Circular Current Loop	Magnetic Field Due to a Current Element

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CBSE CLASS XII Related Questions

1.
A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:
- 80 Wb
- 16 Wb
- $8 \times 10^{-2}$ Wb
- $8 \times 10^{-3}$ Wb
View Solution
2.
Determine the current in the $ 3 \, \Omega $ branch of a Wheatstone Bridge in the circuit shown in the figure.
View Solution
3.
Consider a cylindrical conductor of length $l$ and area of cross-section $A$. Current $I$ is maintained in the conductor and electrons drift with velocity $\vec{v}_d$ ($|\vec{v}_d| = \frac{e|\vec{E}|}{m}\tau$), (where symbols have their usual meanings). Show that the conductivity $\sigma$ of the material of the conductor is given by
$$\sigma = \frac{ne^2}{m}\tau.$$
View Solution
4.
Consider a cylindrical conductor of length $l$ and area of cross-section $A$. Current $I$ is maintained in the conductor and electrons drift with velocity $\vec{v}_d$ ($|\vec{v}_d| = \frac{e|\vec{E}|}{m}\tau$), (where symbols have their usual meanings). Show that the conductivity $\sigma$ of the material of the conductor is given by
$$\sigma = \frac{ne^2}{m}\tau.$$
View Solution
5.
A current of $4.0\,\text{A}$ flows through a wire of length $1\,\text{m}$ and cross-sectional area $1.0\,\text{mm}^2$, when a potential difference of $2\,\text{V}$ is applied across its ends. Calculate the resistivity of the material of the wire.
View Solution
6.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.
View Solution