Why Does Reversing Current Source in EMF Problems Work?

In summary, the conversation is about a homework problem involving a coil and a circular ring with a common axis. The solution involves using the concept of mutual inductance, which makes the problem simpler. The summary also includes the EMF induced in the ring and the magnetic field due to current in the ring.
  • #1
joker_900
64
0
First up, I have got the answer required - however I don't know why this has produced the right answer. I thought it might and tried it, but I don't know why it worked! I would appreciate some insight.

Homework Statement


A small coil of N turns and area A carrying a constant current I and a circular ring with radius R have a common axis. The small coil moves along the axis so that it's distance from the centre of the ring is given by d = d0 + acoswt. Show that the EMF induced in the ring is

(3/2) mu NAIw aR^2 d (R^2 + d^2)^(-5/2) sinwt


Homework Equations





The Attempt at a Solution



OK so for some reason, it worked when I switched the problem around - I said the current is in the ring, not the coil. Then from a previous problem I knew the magnetic field due to current in the ring at any point on its axis is in the direction of the axis and magnitude

0.5 mu IR^2 (R^2 + x^2)^(3/2)

As the coil is small, the field can be considered constant across it's cross-section, and along it's length, so each turn has the same uniform field through it and the total flux linkage is

0.5 Nmu IR^2 (R^2 + d^2)^(3/2)

Then I differentiated to get the EMF and the answer comes out.

But why am I allowed to pretend the current is in the ring instead of the coil? Is it something to do with mutual inductance?

Thanks
 
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  • #2
joker_900 said:
Is it something to do with mutual inductance?

Spot on! This is one of those problems where a direct solution would be miserable, but become very simple when invoking the argument of mutual inductance.
 
  • #3


Dear student,

Thank you for reaching out for clarification on your solution. It is great that you were able to come up with the correct answer, even if you are unsure of why it works. Let me try to provide some insight into your approach.

First, it is important to understand that the magnetic field around a current-carrying wire is circular and perpendicular to the wire. This means that the magnetic field at any point on the axis of the ring will also be circular and perpendicular to the axis. This is a key concept to keep in mind.

Now, let's look at your solution. You correctly identified that the magnetic field at any point on the axis of the ring is given by:

B = 0.5 mu IR^2 (R^2 + x^2)^(3/2)

This is the field due to a current-carrying ring, and it is in the direction of the axis of the ring. This is important because it means that the magnetic field is perpendicular to the direction of motion of the coil. This is what allows us to use Faraday's law to calculate the induced EMF.

In your solution, you assumed that the current is in the ring instead of the coil. This is a valid assumption because the magnetic field at any point on the axis of the ring is the same, regardless of whether the current is in the ring or the coil. This is due to the fact that the magnetic field at any point on the axis of the ring is determined by the current in the entire ring, not just a specific section of the ring.

So, by assuming that the current is in the ring, you were able to use Faraday's law to calculate the induced EMF. This is because the changing magnetic field due to the current in the ring will induce an EMF in the coil, which is moving along the axis of the ring.

In summary, your approach is valid because the magnetic field on the axis of the ring is determined by the current in the entire ring, not just a specific section of the ring. This allows us to use Faraday's law to calculate the induced EMF, even if the current is in the ring instead of the coil.

I hope this helps to clarify your solution. Keep up the good work in your studies!

Sincerely,
 

1. What is a magnetic field?

A magnetic field is an invisible force field that is created by electrically charged particles in motion. It is responsible for the attraction and repulsion of magnetic materials and can also affect the motion of charged particles.

2. How is a magnetic field created?

A magnetic field is created by moving electric charges, such as electrons, in a specific direction. This can occur naturally, such as in the Earth's core, or artificially through the use of magnets or electric currents.

3. What is the unit of measurement for magnetic field?

The unit of measurement for magnetic field is the tesla (T). It is named after the inventor Nikola Tesla and is equivalent to one newton per ampere-meter.

4. How does a magnetic field affect objects?

A magnetic field can cause objects with magnetic properties to move or align themselves in a certain direction, depending on the strength and direction of the field. It can also induce electric currents in conductive materials.

5. How is a magnetic field used in everyday life?

Magnetic fields have numerous applications in everyday life, such as in the production of electricity, transportation (e.g. trains and motors), medical imaging (e.g. MRI machines), and data storage (e.g. hard drives). They are also used in household items like speakers, televisions, and credit cards.

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