A parallel-plate capacitor, made of two circular plates of radius R = 10 cm, is connected in series with a resistor of resistance R = 100 Ω and a battery of emf E = 10 V. What is the maximum value of the magnetic field induced by the displacement current between the plates, at a distance r = 5 cm from the axis of the plates while the capacitor is being charged?

Explanation:

At the time when the capacitor starts charging , current through will be maximum .

Current in the circuit = 10 / 100 = .1 A .

This is called displacement current Id = .1 A .

magnetic field between the plates will be due to this displacement current .

Magnetic field lines  will be  circular in shape.

If B be the magnetic field , we shall apply ampere's circuital law to find magnetic field due to displacement current at a distance of 5 cm from axis.

∫ B dl = μ₀ Id

B x 2π x  .05  = 8.85 x 10⁻¹² x .1

B = 2.82 x 10⁻¹² T .

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.