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 μ0 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/ (Aε0d/d))where A =area of the capacitor.
• E=Q/(Aε0)
• Direction of the electric field will be perpendicular to the selected surface i.e. if considering plate of the capacitor as surface.
• As E =0 outside the plates and E=(Q/(Aε0)) 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=ΦE= (EA) =(QA)/ (Aε0) =(Q/ ε0)
• Assuming Q (charge on capacitor i.e. charging or discharging of the capacitor) changes with time current will be get generated.
• Therefore current Id =(dQ/dt)
• Where Id =displacement current
• =>DifferentiatingΦE =(Q/ ε0) on both sides w.r.t time,
• (dΦE/dt) =(1/ ε0) (dQ/dt)
• where (dQ/dt) =current
• Therefore (dQ/dt) = ε0 (d ΦE/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.
• Id = (dQ/dt) = Displacement current
• Therefore Change in electric field gave rise to Displacement current.
• Current won’t be 0 it will be Id.
• 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.
• 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. 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–1.

(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,

XC = 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 = (μ0r)/ (2 R2) I0 Where,

μ0 = Free space permeability = 4x πx10-7NA-2

I0 = Maximum value of current =√2 I

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

Therefore B = (4xπ x10-7x0.03x√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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