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.

- 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.

- 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/(√ μ**_{0}ε_{0})

- Where μ
_{0}(permeability) and ε_{0}(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

** Answer:- ** The time constant of the CR circuit is τ = CR = 10

= 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)/ (ε_{0}A)

=q/ (πε_{0}); A = π (1)^{2} m^{2} = 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 Φ_{E} through this loop is

Φ_{E} = E × area of the loop

=Exπ x (1/2)^{2}

= (π E/4)

= (q/4ε_{0})

The displacement current

I_{d} =ε_{0}(dΦ_{E}/dt)

= (1/4) (dq/dt)

=0.5x10^{-6} exp (-1)

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

=B x 2π x (1/2)

=μ_{0}(I_{C}+I_{d})

=μ_{0}(0 +I_{d})

=0.5x10^{-6}μ_{0}exp (-1)

or B = 0.74 × 10^{–13} T

.