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.
    A 500 nm photon is incident normally on a perfectly reflecting surface and is reflected. The value of momentum transferred to the surface is:

      • \( 3.87 \times 10^{-43} \, \text{kg} \, \text{ms}^{-1} \)
      • \( 2.5 \times 10^{-30} \, \text{kg} \, \text{ms}^{-1} \)
      • \( 2.65 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \)
      • \( 1.33 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \)

    • 2.
      A circular coil of 100 turns and radius \( \left(\frac{10}{\sqrt{\pi}}\right) \, \text{cm}\) carrying current of \( 5.0 \, \text{A} \) is suspended vertically in a uniform horizontal magnetic field of \( 2.0 \, \text{T} \). The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate:
      the magnetic dipole moment of the coil, and
      the magnitude of the counter torque that must be applied to prevent the coil from turning.


        • 3.
          Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))


            • 4.
              Determine the current in the \( 3 \, \Omega \) branch of a Wheatstone Bridge in the circuit shown in the figure.


                • 5.
                  The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

                    • \( 0.5 \times 10^3 \, \text{m}^{-1} \)
                    • \( 6.0 \times 10^2 \, \text{m}^{-1} \)
                    • \( 7.5 \times 10^2 \, \text{m}^{-1} \)
                    • \( 1.5 \times 10^3 \, \text{m}^{-1} \)

                  • 6.
                    A part of a wire carrying \( 2.0 \, \text{A} \) current and bent at \( 90^\circ \) at two points is placed in a region of uniform magnetic field \( \vec{B} = -0.50 \, \hat{k} \, \text{T} \), as shown in the figure. Calculate the magnitude of the net force acting on the wire.

                      CBSE CLASS XII Previous Year Papers

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