Question

As a parallel-plate capacitor with circular plates 28 cm in diameter is being charged, the current density of the displacement current in the region between the plates is uniform and has a magnitude of 20 A/m2.(a)Calculate the magnitudeBof the magnetic field at a distancer= 69 mm from the axis of symmetry of this region.(b)CalculatedE/dtin this region.

Answer 1

Answer:

magnetic field = 8.67 ×$10^{-7}$ T

dE/dt = 2.25×$10^{-12}$ V/ms

Explanation:

given data

diameter = 28 cm

current = 20 A/m²

r= 69 mm = 69 ×$10^{-3}$  m

to find out

magnetic field and dE/dt

solution

we will apply here magnetic field formula that is

magnetic field = μ I / 2πr    .............1

put here all value

magnetic field = μ (Jπr²) / 2πr

magnetic field = 4π ×$10^{-7}$ (20) ( 69 ×$10^{-3}$) / 2

magnetic field = 8.67 ×$10^{-7}$ T

and dE/dt is calculated by current  that is

current I = ∈A dE/dt

so

dE/dt = I / ∈A      ....................2

that is dE/dt = J / ∈

dE/dt = 20 / 8.85 ×$10^{-12}$

dE/dt = 2.25×$10^{-12}$ V/ms