Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Consider a parallel plate capacitor. The plates arecircular (and obviously made

ID: 1501980 • Letter: C

Question

Consider a parallel plate capacitor. The plates arecircular (and obviously made of conducting material). The plates have a cross-sectional area of 6.07 cm2 and a plate separation of 3.18 mm. The medium between the plates is vacuum and the capacitor is uncharged.

Now, from a certain time onwards (call it t=0 ), a constant charging/conduction current iC ofconstant value 2.30 mA is provided to this capacitor to charge the plates. [You are not told of whether it is a RC or RL circuit or what and the lack of that information should not matter].

Question : Calculate the magnitude of the Magnetic field (if any) inside/within the region between the plates of the capacitor at a radial distance of 0.87 cm from an imaginary line joining the centers of the plates (like an axis). NOTE: The unit has a factor. Be sure to account for that factor BEFORE entering your answer.

please show all steps and deriviations of equations used. thank you!

Explanation / Answer

let R is the radius of plates.

Apply, A = pi*R^2

R = sqrt(A/pi)

= sqrt(6.07*10^-4/pi)

= 0.014 m

current through capacitor, I = 2.3 mA = 2.3*10^-3 A

at r = 0.87 cm = 0.0087 m

Apply, B = mue*I*r/(2*pi*R^2)

= 4*pi*10^-7*2.3*10^-3*0.0087/(2*pi*0.014^2)

= 2.04*10^-8 T

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote