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

magnetic field = 8.67 ×
`10^(-7)` T

dE/dt = 2.25×
`10^(-12)` V/ms

Step-by-step 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