Test III -Part I Due Thursday April 14\" 4:00 pm Prompt Note: Show formulas and

Question

Test III -Part I Due Thursday April 14" 4:00 pm Prompt Note: Show formulas and all work and reasoning on separate answer sheets to get any form of credit. Leaving out vital steps and reasoning will result in point deduction even if your final answer is correct alt in point deduction 1. (20 points) The planar circular plates of a capacitor are being charged. At a given moment the charge is being built up at the rate separation d The plates have a radius R and a Show that the magnetic field due to the displacement current midway between the plates at a radius equal to half the plate radius is given by a. Holdq 4m R de Given that the radius of the plate is R = 0.1 m, and the plate separation d = lem. Calculate the magnetic field due to the displacement current midway between the plates at a radius equal to half the plate radius, if the charge builds up at a rate of 1 Cs. b.

Explanation / Answer

a) From Ampere’s Law,

Integration(B.dl) = uo*(I + ID)

Where ID = Displacement current

When r < R      where R = Radius of plates

B*2pi*r = uo*(I/pi*R^2)*pi*r^2

B = (uo*I*r)/(2pi*R^2)

For r = R/2,    B = (uo*I*R/2)/(2pi*R^2) = (uo/4pi*R)*(dq/dt)

b) B = (10^-7/0.1)*1 = 10^-6 T

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.