Show what the magnitude of induced emf

Click For Summary
SUMMARY

The discussion focuses on calculating the magnitude of induced electromotive force (emf) in a magnetic field defined as B = K(x³z², 0, -x²z³)sin(ωt). The derived formula for the induced emf around a circle R in the plane z = a is ε = (K/4)πa³R⁴ωcos(ωt). Key steps include integrating the magnetic field component Bz and converting to polar coordinates for proper evaluation. Participants emphasize the importance of correctly applying trigonometric identities and understanding variable dependencies in double integrals.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically Faraday's law of induction.
  • Familiarity with vector calculus, particularly surface integrals.
  • Knowledge of polar and cylindrical coordinates for integration.
  • Proficiency in trigonometric identities, especially cos(2θ) = 2cos²(θ) - 1.
NEXT STEPS
  • Study the derivation of Faraday's law of induction in various coordinate systems.
  • Learn about the application of surface integrals in electromagnetism.
  • Explore the use of cylindrical coordinates in solving electromagnetic problems.
  • Review trigonometric identities and their applications in integration techniques.
USEFUL FOR

Students and professionals in physics, electrical engineering, and applied mathematics, particularly those working with electromagnetic fields and induction phenomena.

auk411
Messages
54
Reaction score
0
Someone who knows what they are talking about: Show what the magnitude of induced emf

Homework Statement



Consider a magnetic field B = K(x3z2,0, -x2z3)sinωt in the region of interest, where K and ω are positive constants and t is variable time. Show that the magnitude of the induced emf around a circle R in the plane z = a with its center at x = 0, y = 0, z = a is:
ε = (K/4)∏a3R4ωcosωt

Homework Equations



Fluxb = ∫B . dA

The Attempt at a Solution



Since the normal vector points in the k direction, we only have to worry about Bz.

∫Bzdydx. So -∫∫(sinwt)x2a3dydx.

The make the change to polar:

-aK3∫∫(sinwt)(rcosθ)2r dr dθ = -(K/4)a3R4∫cosθsin(wt) dθ.

This doesn't get me anywhere. I'm not really sure what I'm supposed to be integrating over, which is probably why I'm stuck.
 
Last edited:
Physics news on Phys.org
Hi auk411! :smile:

Let's start with Bz.
Can you write it down separately?
It seems you did not copy it correctly.

Secondly you did not bring the constant out of the integral properly.

Furthermore in polar coordinates you would integrate r from 0 to R, and theta from 0 to 2pi.

And for the tip: rewrite (cosθ)2 using cos2θ.
 
I like Serena said:
Hi auk411! :smile:

Let's start with Bz.
Can you write it down separately?
It seems you did not copy it correctly.

Secondly you did not bring the constant out of the integral properly.

Furthermore in polar coordinates you would integrate r from 0 to R, and theta from 0 to 2pi.

And for the tip: rewrite (cosθ)2 using cos2θ.

(cosθ)2 using cos2θ.[/QUOTE] ... huh, what trig identity are you using.

this still doesn't answer the most pressing question. t varies, theta varies and r varies. we have 3 varying variables in a DOUBLE integral. I see no way to reduce them to two. How do I get around this.
 
First things first.
You appear to have skipped my question, so I'll answer it myself:

Bz = -Kx2a3sinωt

This is not what you used.

auk411 said:
(cosθ)2 using cos2θ. ... huh, what trig identity are you using.

cos(2θ) = 2 cos2θ - 1


auk411 said:
this still doesn't answer the most pressing question. t varies, theta varies and r varies. we have 3 varying variables in a DOUBLE integral. I see no way to reduce them to two. How do I get around this.

No, you have 2 variables.
You appear to be thinking spherical coordinates, but you should be thinking in cylindrical coordinates.

z is constant at z=a.
Only the other 2 vary.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 7 ·
Replies
7
Views
5K