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.

magnetic field = 8.67 × T

dE/dt = 2.25× V/ms

Explanation:

given data

diameter = 28 cm

current = 20 A/m²

r= 69 mm = 69 ×  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 × (20) ( 69 ×) / 2

magnetic field = 8.67 × 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 ×

dE/dt = 2.25× V/ms

found in the nucleus

has mass of one amu

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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