Class 12 Physics Electromagnetic Waves | Maxwells correction to Amperes law |

**Maxwell’s correction to Ampere’s law**

- Ampere’s law states that “the line integral of resultant magnetic field along a closed plane curve is equal to μ
_{0}time the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant". - Ampere’s law is true only for steady currents.
- Maxwell found the shortcoming in Ampere’s law and he modified Ampere’s law to include time-varying electric fields.
- For Ampere’s circuital law to be correct Maxwell assumed that there hasto be some current existing between the plates of the capacitor.
- Outside the capacitor current was due to the flow of electrons.
- There was no conduction of charges between the plates of the capacitor.
- According to Maxwell between the plates of the capacitor there is an electric field which is directed from positive plate to the negative plate.
- Magnitude of the electric field E =(V/d)
- Where V=potential difference between the plates, d = distance between the plates.

- E = (Q/Cd)
- where Q=charge on the plates of the capacitor,Capacitance of the capacitor=C

- =>= (Q/ (Aε
_{0}d/d))where A =area of the capacitor. **E=Q/(Aε**_{0})- Direction of the electric field will be perpendicular to the selected surface i.e. if considering plate of the capacitor as surface.

- Magnitude of the electric field E =(V/d)
- As E =0 outside the plates and E=(Q/(Aε
_{0})) between the plates.- There may be some electric field between the plates because of which some current is present between the plates of the capacitor.
- Electric Flux through the surface=Φ
_{E}= (EA) =(QA)/ (Aε_{0}) =(Q/ ε_{0})

- Assuming Q (charge on capacitor i.e. charging or discharging of the capacitor) changes with time current will be get generated.
- Therefore current I
_{d}=(dQ/dt)- Where I
_{d}=displacement current

- Where I
- =>DifferentiatingΦ
_{E}=(Q/ ε_{0}) on both sides w.r.t time, - (dΦ
_{E}/dt) =(1/ ε_{0}) (dQ/dt)- where (dQ/dt) =current

- Therefore (dQ/dt) = ε
_{0 }(d Φ_{E}/dt) - =>Current was generated because of change of electric flux with time.
- Electric flux arose because of presence of electric field in the plates of the capacitor.
**I**_{d}= (dQ/dt) = Displacement current- Therefore Change in electric field gave rise to Displacement current.
- Current won’t be 0 it will be I
_{d}. - There is some current between the plates of the capacitor and there is some current at the surface.
- At certain points there is no displacement current there is only conduction current and vice-versa.

- Current won’t be 0 it will be I
- Maxwell corrected the Ampere’s circuital law by including displacement current.
- He said that there is not only the current existed outside the capacitor but also current known as displacement currentexisted between the plates of the capacitor.
- Displacement current exists due to the change in the electric field between the plates of the capacitor.
**Conclusion**:-Magnetic fields are produced both by conduction currents and by time varying fields.

- Therefore current I

** Problem:- **A parallel plate capacitor (Fig) made of circular plates each of radiusR = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected toa 230 V ac supply with a (angular) frequency of 300 rad s

(a) What is the rms value of the conduction current?

(b) Is the conduction current equal to the displacement current?

(c) Determine the amplitude of B at a point 3.0 cm from the axisbetween the plates.

** Answer:- **Radius of each circular plate, R = 6.0 cm = 0.06 m

Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10^{−12} F

Supply voltage, V = 230 V

Angular frequency, ω = 300 rad s^{−1}

(a) Rms value of conduction current, I

Where,

X_{C} = Capacitive reactance

=1/ (ωC)

Therefore, I = V × ωC

= 230 × 300 × 100 × 10^{−12}

= 6.9 × 10^{−6} A

= 6.9 μA

Hence, the rms value of conduction current is 6.9 μA.

(b) Yes, conduction current is equal to displacement current.

(c) Magnetic field is given as:

B = (μ_{0}r)/ (2 R^{2}) I_{0} Where,

μ_{0} = Free space permeability = 4x πx10^{-7}NA^{-2}

I_{0} = Maximum value of current =√2 I

r = Distance between the plates from the axis = 3.0 cm = 0.03 m

Therefore B = (4xπ x10^{-7}x0.03x√2x6.9x10^{-6})/ (2xπ x0.06)^{2}

= 1.63 × 10^{−11} T

Hence, the magnetic field at that point is 1.63 × 10^{−11} T.

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