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

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:

Answers and Replies

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kuruman
Science Advisor
Homework Helper
Gold Member
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.

Zack K
Here area should be taken of stationary loop because changing flux link to this area

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.

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.

kuruman
rude man
Homework Helper
Gold Member
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
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.