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

AI Thread Summary
The discussion focuses on calculating the magnetic flux of a moving coil and its effect on a stationary coil. The magnetic flux is derived using the formula involving the magnetic field and area, leading to an expression that incorporates the distance term, z, which raises questions about its relevance to the stationary coil. Participants clarify that the computed flux for the moving coil can be used to determine the electromotive force (emf) in the stationary coil, assuming uniform flux around both coils. It is noted that while the situation is more complex due to mutual and self-inductance, the analysis simplifies by treating the flux as uniform. The conversation emphasizes the importance of understanding the relationship between the moving coil's flux and the stationary coil's emf.
Zack K
Messages
166
Reaction score
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
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
Here area should be taken of stationary loop because changing flux link to this area
 
  • Like
Likes Miles123K
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.
 
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
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top