Capacitor Field And Distance-Electricity Solved Problem, Exercises of Electrical Engineering

This is solution to problems related Electrical Engineering course. It was suggested by Prof. Bhooshan Sawhney at Shree Ram Swarup College of Engineering and Management. It includes: Radius, Capacitor, Distance, Axis, Points, Magnitude, Magnetic, Field, Maximum, Result

Typology: Exercises

2011/2012

Uploaded on 07/20/2012

apurva
apurva 🇮🇳

70 documents

1 / 1

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
26. Let Rbe the radius of a capacitor plate and rbe the distance from axis of the capacitor. For points
with rR, the magnitude of the magnetic field is given by
B=µ0ε0r
2
dE
dt ,
and for rR,itis
B=µ0ε0R2
2r
dE
dt .
The maximum magnetic field occurs at points for which r=R, and its value is given by either of the
formulas above:
Bmax =µ0ε0R
2
dE
dt .
There are two values of rfor which B=Bmax/2: one less than Rand one greater. To find the one that
is less than R,wesolve µ0ε0r
2
dE
dt =µ0ε0R
4
dE
dt
for r. The result is r=R/2=(55.0mm)/2=27.5 mm. To find the one that is greater than R,wesolve
µ0ε0R2
2r
dE
dt =µ0ε0R
4
dE
dt
for r. The result is r=2R=2(55.0 mm) = 110 mm.
docsity.com

Partial preview of the text

Download Capacitor Field And Distance-Electricity Solved Problem and more Exercises Electrical Engineering in PDF only on Docsity!

  1. Let R be the radius of a capacitor plate and r be the distance from axis of the capacitor. For points with r ≤ R, the magnitude of the magnetic field is given by

B = μ^0 ε^0 r 2

dE dt

and for r ≥ R, it is B = μ 0 ε 0 R^2 2 r

dE dt

The maximum magnetic field occurs at points for which r = R, and its value is given by either of the formulas above: Bmax = μ 0 ε 0 R 2

dE dt

There are two values of r for which B = Bmax/2:one less than R and one greater. To find the one that is less than R, we solve μ 0 ε 0 r 2

dE dt

μ 0 ε 0 R 4

dE dt for r. The result is r = R/2 = (55.0 mm)/2 = 27.5 mm. To find the one that is greater than R, we solve

μ 0 ε 0 R^2 2 r

dE dt =^

μ 0 ε 0 R 4

dE dt for r. The result is r = 2R = 2(55.0 mm) = 110 mm.

docsity.com