/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q12P Question: A parallel-plate capac... [FREE SOLUTION] | 91影视

91影视

Question: A parallel-plate capacitor with circular plates of radius 40 mm is being discharged by a current of 6.0 A . At what radius (a) inside and (b) outside the capacitor, the gap is the magnitude of the induced magnetic field equal to 75% of its maximum value? (c) What is that maximum value?

Short Answer

Expert verified

Answer

  1. The radius inside the capacitor gap at which the magnitude of the induced magnetic field is equal to 75% of its maximum value is r1=30mm
  2. The radius outside the capacitor gap at which the magnitude of the induced magnetic field is equal to 75% of its maximum value is r2=53mm
  3. The maximum value of magnetic field Bmax=3.010-5T

Step by step solution

01

Step 1: Given information

The radius of a plate of parallel plate capacitor is,r=40mm ,

Discharged current is,i=6.0A .

02

Understanding the concept  

The magnetic field inside a capacitor is directly proportional to the loop radius The relation is written as below:

B=(0id2R2)r (i)

Here, is the magnetic field, 0is permeability constant, i is current, R is the inside radius, and r is the outside radius.

The magnetic field outside the capacitor is inversely proportional to the loop radius. The relation is written as below,

B=(0id2) (ii)

Here, B is the magnetic field, 0is permeability constant, i is current, and r is the outside radius.

03

(a) Determining at what radius inside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value

From equation 32-16, the magnetic field inside the capacitor is directly proportional to the radius so that, Bmaxoccurs when the r =R .

BmaxR, and, the given condition is 0.75Bmaxr1.

So, to find the value of r1 by taking the ratio of these two equations,

0.75BmaxBmax=r1Rr1=0.75R

By substituting the value, the result is,

r1=0.7540=30mm

Hence, the radius inside the capacitor gap at which the magnitude of the induced magnetic field is equal to 75% of its maximum value is r1=30mm

04

(b) Determining at what radius outside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value 

Similarly, for outside the capacitor, Bmax occurs, when r =R , and from equation 32- the magnetic field is inversely proportional to . So, the maximum magnetic field is at r =R . From the given condition, 0.75Bmax occurs when the radius is r2, so according to equation 32-17, write proportionality as,

0.75Bmax1r2

(iii)

Also, the magnetic field is inversely proportional to the inside radius, therefore,

Bmax1R (iv)

Taking a ratio of equations (iii) and (iv), we get

0.75BmaxBmax=Rr2

Solve the above equation to calculate as,

r2=R0.75=40mm0.75=53mm

Hence, the radius outside the capacitor gap at which the magnitude of the induced magnetic field is equal to 75% of its maximum value isr2=53mm .

05

(c) Determining the maximum value of the magnetic field 

Bmax can be calculated using equation (ii) as,

Bmax=410-7Tm/A6.0A24010-3m=3.010-5TBmax=3.010-5T

Hence, the maximum value of the magnetic field =3.010-5T.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The figure 32-20 shows a circular region of radiusR=3cm in which a displacement currentis directedout of the page. The magnitude of the density of this displacement current is Jd=(4A/m2)(1-r/R), where is the radial distance (rR).(a) What is the magnitude of the magnetic field due to displacement current at 2cm?(b)What is the magnitude of the magnetic field due to displacement current at5cm ?

The magnetic flux through each of five faces of a die (singular of 鈥渄ice鈥) is given by B=NWb, where N(= 1 to 5) is the number of spots on the face. The flux is positive (outward) for Neven and negative (inward) for Nodd. What is the flux through the sixth face of the die?

A capacitor with square plates of edge length L is being discharged by a current of 0.75 A. Figure 32-29 is a head-on view of one of the plates from inside the capacitor. A dashed rectangular path is shown. If L = 12cm, W = 4.0 cm , and H = 2.0 cm , what is the value B鈬赌ds鈬赌of around the dashed path?

A capacitor with parallel circular plates of the radius R=1.20cmis discharging via a current of 12.0 A . Consider a loop of radiusR/3that is centered on the central axis between the plates. (a)How much displacement current is encircled by the loop? The maximum induced magnetic field has a magnitude of 12.0 mT. (b)At what radius inside and (c)outside the capacitor gap is the magnitude of the induced magnetic field 3.00 mT?

A parallel-plate capacitor with circular plates of the radius Ris being charged. Show that, the magnitude of the current density of the displacement current is

role="math" localid="1663149858775" Jd=0(dEdt)forrR.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.