Current induced on a coil by a changing magnetic flux in another coil

In summary,The moving coil's flux is a function of dz/dt which is also the flux seen by the stationary coil. So just compute the emf developed in the stationary coil from that and then the current in the stationary coil.
  • #1
Zack K
166
6
Homework Statement
A thin coil is located at the origin; its radius is 5 cm and its axis lies on the##x##axis. It has 40 low-resistance turns and is connected to a 80 Ω resistor. A second thin coil is located at <15, 0, 0> cm and is traveling toward the origin with a speed of 8 m/s; it has 50 low-resistance turns, its axis lies on the axis, its radius is 4 cm, and it has a current of 17 A, powered by a battery. What is the magnitude of the current in the first coil?

EDIT: I managed to get the answer, but I'm not sure where the "mark solved" button is.
Relevant Equations
##emf = -\frac {d\phi}{dt}##
##\frac {d\phi}{dt} = -\frac {dB}{dt}A##(A is the cross sectional area of the loop)
I will first calculate the magnetic flux of the coil in motion.

$$\frac {d\phi}{dt} = -\frac {dB_{loop}}{dt}A = -\frac{d}{dt}(\frac{\mu_o}{4\pi}\frac {2\pi NR^2I}{(R^2+z^2)^{\frac{3}{2}}})A$$differentiating in terms of ##z##, we get $$\frac {d\phi}{dt} =(\frac{\mu_o}{4\pi}\frac {6\pi^2 NR^4Iz}{(R^2+z^2)^{\frac{5}{2}}})\frac{dz}{dt}$$The area was just ##\pi R^2## and I just brought that into the expression. ##\frac{dz}{dt}## is our velocity.
Assuming this process is correct, I'm not sure on how to find the emf on the stationary coil. Is this expression for the coil in motion, or for the stationary coil? The distance term ##z## is making me think it's for the stationary coil, because why would a coils own change in flux have a distance term for it?
 
Last edited:
Physics news on Phys.org
  • #2
I wouldn't worry about the "Mark solved" button. PF is being upgraded as we speak. The button will eventually appear (I think) when the upgrade is completed.
 
  • Like
Likes Zack K
  • #3
Here area should be taken of stationary loop because changing flux link to this area
 
  • Like
Likes Miles123K
  • #4
Take ##A## as the area of the stationary loop and you'll be fine. I am not sure if assuming the magnetic field in the stationary loop is uniform is a reasonable assumption.
 
  • #5
kuruman said:
I wouldn't worry about the "Mark solved" button. PF is being upgraded as we speak. The button will eventually appear (I think) when the upgrade is completed.
At the moment I don't have short term plans to bring it back.
 
  • Informative
Likes kuruman
  • #6
Zack K said:
I will first calculate the magnetic flux of the coil in motion.

$$\frac {d\phi}{dt} = -\frac {dB_{loop}}{dt}A = -\frac{d}{dt}(\frac{\mu_o}{4\pi}\frac {2\pi NR^2I}{(R^2+z^2)^{\frac{3}{2}}})A$$differentiating in terms of ##z##, we get $$\frac {d\phi}{dt} =(\frac{\mu_o}{4\pi}\frac {6\pi^2 NR^4Iz}{(R^2+z^2)^{\frac{5}{2}}})\frac{dz}{dt}$$The area was just ##\pi R^2## and I just brought that into the expression. ##\frac{dz}{dt}## is our velocity.
Assuming this process is correct, I'm not sure on how to find the emf on the stationary coil. Is this expression for the coil in motion, or for the stationary coil? The distance term ##z## is making me think it's for the stationary coil, because why would a coils own change in flux have a distance term for it?
You've computed the B field
Zack K said:
I will first calculate the magnetic flux of the coil in motion.

$$\frac {d\phi}{dt} = -\frac {dB_{loop}}{dt}A = -\frac{d}{dt}(\frac{\mu_o}{4\pi}\frac {2\pi NR^2I}{(R^2+z^2)^{\frac{3}{2}}})A$$differentiating in terms of ##z##, we get $$\frac {d\phi}{dt} =(\frac{\mu_o}{4\pi}\frac {6\pi^2 NR^4Iz}{(R^2+z^2)^{\frac{5}{2}}})\frac{dz}{dt}$$The area was just ##\pi R^2## and I just brought that into the expression. ##\frac{dz}{dt}## is our velocity.
Assuming this process is correct, I'm not sure on how to find the emf on the stationary coil. Is this expression for the coil in motion, or for the stationary coil? The distance term ##z## is making me think it's for the stationary coil, because why would a coils own change in flux have a distance term for it?
You've computed the moving coil's flux as a function of dz/dt which is also the flux seen by the stationary coil. So just compute the emf developed in the stationary coil from that and then the current in the stationary coil.
Yes, you have to assume the flux is uniform in & around both coils. Computing the actual flux distributions is prohibitively difficult.

It's actually quite a bit more complicated than that; in reality there is also mutual and self-inductance. But you are to ignore all that and just assume the flux in the stationary coil is the computed flux a distance z away from the moving coil.
 

1. How does a changing magnetic flux induce a current in a coil?

When a magnetic field changes, it creates a changing flux, which in turn induces an electric field. This electric field then causes a current to flow in a nearby conductor, such as a coil.

2. What is the relationship between the strength of the magnetic field and the induced current?

The strength of the induced current is directly proportional to the strength of the changing magnetic field. This means that the stronger the magnetic field, the greater the induced current will be.

3. Why is the direction of the induced current important?

The direction of the induced current is important because it follows Lenz's law, which states that the direction of the induced current will always oppose the change in the magnetic field that caused it. This ensures that energy is conserved in the system.

4. Can a coil induce a current in itself?

Yes, a coil can induce a current in itself if there is a changing magnetic field passing through it. This is known as self-induction and occurs when the magnetic field produced by the current in the coil changes, inducing a current in the same coil.

5. How can the induced current in a coil be increased?

The induced current in a coil can be increased by increasing the rate of change of the magnetic field, increasing the number of turns in the coil, or increasing the strength of the magnetic field. Additionally, using a ferromagnetic core inside the coil can also amplify the induced current.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
520
  • Introductory Physics Homework Help
Replies
3
Views
171
  • Introductory Physics Homework Help
Replies
7
Views
81
  • Introductory Physics Homework Help
Replies
3
Views
153
  • Introductory Physics Homework Help
Replies
7
Views
308
  • Introductory Physics Homework Help
Replies
2
Views
808
  • Introductory Physics Homework Help
Replies
12
Views
208
  • Introductory Physics Homework Help
Replies
1
Views
160
  • Introductory Physics Homework Help
Replies
7
Views
862
Back
Top