Class 12 Physics Electromagnetic Waves Maxwells Equations

Maxwell’s Equations

  • Maxwell's equations describe how an electric field can generate a magnetic field and vice-versa. These equations describe the relationship and behaviour of electric and magnetic fields.
  • Maxwell gave a set of 4 equations which are known as Maxwell’s equations.
  • According to Maxwell equations:-
    • A flow of electric current will generate magnetic field and if the current varies with time magnetic field will also give rise to an electric filed.
      • First equation (1) describes the surface integral of electric field.
      • Second equation (2) describes the surface integral of magnetic field.
      • Third equation (3) describes the line integral of electric field.
      • Fourth equation (4) describes line integral of magnetic field.


  • Maxwell was the first to determine the speed of propagation of EM waves is same as the speed of light.
  • Experimentally it was found that:-
    • c =1/(√ μ0 ε0)
  • Where μ0(permeability) and ε0(permittivity) and c= velocity of light.
  • Maxwell’s equations show that the electricity, magnetism and ray optics are all inter-related to each other.

Problem:- A parallel plate capacitor with circular plates of radius1 m has a capacitance of 1 nF. At t = 0, it is connected for charging inseries with a resistor R = 1 MΩ across a 2V battery (Fig.). Calculatethe magnetic field at a point P, halfway between the centre and theperiphery of the plates, after t = 10–3 s. (The charge on the capacitorat time t is q (t) = CV [1 – exp (–t/τ)], where the time constant τ isequal to CR.)

Answer:-  The time constant of the CR circuit is τ = CR = 10–3 s. Then,we have

 = q (t) = CV [1 – exp (–t/τ)]

= 2 × 10–9 [1– exp (–t/10–3)]

The electric field in between the plates at time t is

E= q (t)/ (ε0A)

=q/ (πε0); A = π (1)2 m2 = area of the plates.

Consider now a circular loop of radius (1/2) m parallel to the platespassing through P. The magnetic field B at all points on the loop isalong the loop and of the same value.

The flux ΦE through this loop is

ΦE = E × area of the loop

=Exπ x (1/2)2

= (π E/4)

= (q/4ε0)

The displacement current


= (1/4) (dq/dt)

=0.5x10-6 exp (-1)

at t = 10–3s. Now, applying Ampere-Maxwell law to the loop, we get

=B x 2π x (1/2)


0(0 +Id)

=0.5x10-6μ0exp (-1)

or B = 0.74 × 10–13 T

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