Induction and EMF: Homework Statement Solution

In summary, a small ring of radius r is on a common axis and is a hight z above the current carrying ring, where z>>R. The small ring moves up with velocity v. The emf in the upper ring is -dΦ/dt and the direction of the induced current is clockwise.
  • #1
captain.joco
39
0

Homework Statement


In a ring of radius R, there is current I flowing in the counterclockwise direction as viewed from above. A small ring of radius r is on a common axis and is a hight z above the current carrying ring, where z>>R. The small ring moves up with velocity v. Calculate the emf in the upper ring and the direction of the induced current. It can be assumed that the ring is sufficiently small so that the magnetic field across it's area is constant.


Homework Equations





The Attempt at a Solution


So E= - dΦ/dt
Φ = B*A ( since the magnetic field across it's area is constant ).
A = π*r*r
B= (μIR^2) / [ 2 ( R*R + z*z )^3/2 ] but since z>>R we can use the approximation
B= (μIR^2) / (2z^3)

Now i know I need to do the -dΦ/dt but I am getting confused ( how to do it? ).
Also the direction of the induced current will be clockwise. Please help
 
Physics news on Phys.org
  • #2
In order to do the derivative you need to have time involved. The only quantity in that expression for the flux that depends on time is z, correct? Is there any way to get z in terms of time?

HINT: Have you used the velocity, v, yet?
 
  • #3
So will the emf be A* (μIR*R/2) * 3( v^3 * z^4 ) or I did it very very wrongly..
 
  • #4
You should not have z in your final answer. Start with this line:

[tex]\Phi=\frac{\mu IR^2A}{2z^3}[/tex]

Then, remember that: v = z/t.

Can you use the last expression to remove z from the first?
 
  • #5
Sorry I made a typo.. should have been A* (μIR*R/2) * 3/( v^3 * t^4 ) but then still
A* (μIR*R/2) * 3/( z^3*t )
 
  • #6
Is this anywhere near to correct?
 
  • #7
I think you are missing a minus sign in there. You should have picked up a -3 with the differentiation.

Other than that, it looks good.
 
  • #8
But from the original equation which is emf = - dΦ/dt, so this minus cancels out with the -3 minus.. I will redo it just in case. That's for all the help, you can count this one as solved ( I understood the main idea with the z being the only thing that changes with time! ). Thanks very much
 

1. What is induction and how does it relate to EMF?

Induction refers to the production of an electric current in a conductor by changing the magnetic field around it. EMF, or electromagnetic force, is the force that causes this induction to occur. In other words, EMF is the force that drives the flow of electricity in a conductor.

2. What are some real-life applications of induction and EMF?

Some common examples of induction and EMF in everyday life include electric generators, transformers, and induction cooktops. Induction is also used in wireless charging technology, such as in smartphones and electric toothbrushes.

3. How is the strength of an induced current determined?

The strength of an induced current is determined by the rate at which the magnetic field changes and the number of turns in the conductor. This is known as Faraday's law, which states that the induced EMF is directly proportional to the rate of change of the magnetic flux through the conductor.

4. How do you calculate the direction of the induced current?

The direction of the induced current can be determined using Lenz's law, which states that the induced current will flow in such a direction as to oppose the change that caused it. This can be determined by the right-hand rule, where the thumb points in the direction of the changing magnetic field and the curled fingers point in the direction of the induced current.

5. Can induction and EMF be used to generate electricity from renewable sources?

Yes, induction and EMF are integral components in many forms of renewable energy, such as hydroelectric power, wind turbines, and solar panels. These technologies use induction to generate electricity from natural sources, such as water, wind, and sunlight.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
216
  • Introductory Physics Homework Help
Replies
3
Views
282
  • Introductory Physics Homework Help
Replies
1
Views
803
  • Introductory Physics Homework Help
Replies
3
Views
113
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
456
  • Introductory Physics Homework Help
Replies
2
Views
176
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
700
Back
Top