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

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

NCERT SOLUTIONS

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)] = EElectric + Fmagnetic 
  • 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.

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


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CBSE CLASS XII Related Questions

  • 1.
    If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.


      • 2.
        Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

          • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
          • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
          • Assertion (A) is true, but Reason (R) is false.
          • Both Assertion (A) and Reason (R) are false.

        • 3.
          The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

            • \( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
            • \( \frac{q}{\pi \epsilon_0 l^2} \) pointing along AM
            • \( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM
            • Zero

          • 4.
            Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is \( 4 \, \mu\text{F} \). Calculate the potential difference across the plates of X and Y.


              • 5.
                Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.


                  • 6.
                    Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

                      CBSE CLASS XII Previous Year Papers

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