Einstein44
Homework Statement:
When studying Lenzs Law / Faradays Law, I came across two different types of equations and do not know which one is correct. What I need to find is the induced voltage on a copper coil, as a magnet moves through it. Can someone help tell me which one is appropriate to use in this case?
Relevant Equations:
1. E= - ∂ΦB/∂t
where ΦB= rate of change in magnetic flux
E= induced voltage

2. E = -N x Δø/Δt
where N= number of turns in the coil
Δø = change in flux inside the coil

As you can notice the second equation takes into account the number of turns in the coil, whereas the other equation doesn't despite being also used for a coil.
And what is even the difference between having a partial derivative in the equation and instead "delta" for change in magnetic flux over change in time??
.

LeafNinja
Faraday's law is represented in derivative form as
$$\nabla \times \vec{E} = -\frac{d\vec{B}}{dt}$$
In integral form, this is
$$\int_{closedloop} \vec{E} \bullet dl = -\frac{d\Phi_{B}}{dt}$$

Where ##\Phi_{B}## is the magnetic flux or integral of the magnetic flux density over the cross section and ##\int_{closedloop} \vec{E} \bullet dl## is the integral of the electric field around a closed loop.

The - sign indicates that if your are facing the coil and the magnetic field lines are going through the coil pointing towards you, the E-field lines will travel clockwise.

The equations 1. and 2. that you presented use E for the emf which is generally equal to the voltage such as in a battery. Another way to think of it is ##V = \int_{coilpath} \vec{E} \bullet dl ## where ##V## is the voltage.

The N simply indicates that there are N turns of the coil. This is approximated by using Faraday's law on N closed loops. So ##N\int_{closed loop} \vec{E} \bullet dl = E = V = -N\frac{d\Phi_{B}}{dt}## becomes larger (There are N loops or turns, so given a uniform magnetic field, the electromotive force becomes N times as large as it would have been had there only been 1 turn). So the second equation is more correct.

Regarding your question about derivatives, ##\frac{\Delta y}{\Delta t}## indicates finite differences (eg. how far did a frog move in one second).

##\frac{dy}{dt}## indicates derivatives (how far the frog is moving per second at this instant in time)

##\frac{\partial y}{\partial t}## indicates the change in the dependent variable given that only 1 indpendent variable is changing. Eg. If the frog's position depends on both the snake position x and the time t,

$$\frac{dy}{dt} = \frac{\partial y}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial y}{\partial t}$$

So ##\frac{\partial y}{\partial t}## is the rate at which the frog's position would change if the snake was not moving.

In this case either the derivative or partial derivative would work because there is only 1 independent variable: time.

Einstein44
Einstein44
Faraday's law is represented in derivative form as
$$\nabla \times \vec{E} = -\frac{d\vec{B}}{dt}$$
In integral form, this is
$$\int_{closedloop} \vec{E} \bullet dl = -\frac{d\Phi_{B}}{dt}$$

Where ##\Phi_{B}## is the magnetic flux or integral of the magnetic flux density over the cross section and ##\int_{closedloop} \vec{E} \bullet dl## is the integral of the electric field around a closed loop.

The - sign indicates that if your are facing the coil and the magnetic field lines are going through the coil pointing towards you, the E-field lines will travel clockwise.

The equations 1. and 2. that you presented use E for the emf which is generally equal to the voltage such as in a battery. Another way to think of it is ##V = \int_{coilpath} \vec{E} \bullet dl ## where ##V## is the voltage.

The N simply indicates that there are N turns of the coil. This is approximated by using Faraday's law on N closed loops. So ##N\int_{closed loop} \vec{E} \bullet dl = E = V = -N\frac{d\Phi_{B}}{dt}## becomes larger (There are N loops or turns, so given a uniform magnetic field, the electromotive force becomes N times as large as it would have been had there only been 1 turn). So the second equation is more correct.

Regarding your question about derivatives, ##\frac{\Delta y}{\Delta t}## indicates finite differences (eg. how far did a frog move in one second).

##\frac{dy}{dt}## indicates derivatives (how far the frog is moving per second at this instant in time)

##\frac{\partial y}{\partial t}## indicates the change in the dependent variable given that only 1 indpendent variable is changing. Eg. If the frog's position depends on both the snake position x and the time t,

$$\frac{dy}{dt} = \frac{\partial y}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial y}{\partial t}$$

So ##\frac{\partial y}{\partial t}## is the rate at which the frog's position would change if the snake was not moving.

In this case either the derivative or partial derivative would work because there is only 1 independent variable: time.
Thank you. This is really helpful.
Do you know by any chance what the equation for the magnitude of eddy currents is and how I can calculate this using the induced emf? And is there an equation specifically for a solenoid (coil) for the magnitude of eddy currents?

For some reason I was not able to find this anywhere on the internet, so I am interested to know if someone could tell me this.

LeafNinja
I am not sure what you are looking for when you say magnitude of eddy currents. Eddy currents are currents induced by a magnetic field, and flow in loops, usually in sheet conductors rather than wires.

The current in the solenoid can be considered analogous to the flow of water in a pipe: the current is traveling in a desired path, just as water travels along the desired path of the pipe.

In contrast, eddy currents can be compared to eddies in water, which are localized vortices in a body of water.

The equations that describe eddy currents should be derived from Maxwell's equations, as are all electromagnetic phenomenon.

For example, if there was a magnetic field increasing in the direction pointing out of the page towards you, the eddy currents shown in the image would be produced. This is due to Faraday's law, which is the same phenomenon that causes currents in the solenoid. Except, this time, the currents are localized rather than along a certain path and should be thought of as current density ##\vec{J}##

$$\int_{closedloop} \vec{E} \bullet dl = -\frac{d\Phi_{B}}{dt}$$

Because ## \vec{E} = \rho\vec{J}## where ##\rho## is the resistivity, the current density can be obtained if the electric field is known.

Note that the other Maxwell's Equations such as Ampere's law also play a role because the electrical currents themselves give rise to another magnetic field.

Einstein44
Einstein44
I am not sure what you are looking for when you say magnitude of eddy currents. Eddy currents are currents induced by a magnetic field, and flow in loops, usually in sheet conductors rather than wires.

The current in the solenoid can be considered analogous to the flow of water in a pipe: the current is traveling in a desired path, just as water travels along the desired path of the pipe.

In contrast, eddy currents can be compared to eddies in water, which are localized vortices in a body of water.

The equations that describe eddy currents should be derived from Maxwell's equations, as are all electromagnetic phenomenon.

View attachment 286509
For example, if there was a magnetic field increasing in the direction pointing out of the page towards you, the eddy currents shown in the image would be produced. This is due to Faraday's law, which is the same phenomenon that causes currents in the solenoid. Except, this time, the currents are localized rather than along a certain path and should be thought of as current density ##\vec{J}##

$$\int_{closedloop} \vec{E} \bullet dl = -\frac{d\Phi_{B}}{dt}$$

Because ## \vec{E} = \rho\vec{J}## where ##\rho## is the resistivity, the current density can be obtained if the electric field is known.

Note that the other Maxwell's Equations such as Ampere's law also play a role because the electrical currents themselves give rise to another magnetic field.
What I meant was more:
How can I fund the magnetic force of the eddy currents that they apply on the magnet moving through the coil?
I have noticed that this is an incredibly difficult task, as stated by some other members, so my best guess was to do some simplifying assumptions...
Have you got any idea how this can be done?
I was kind of trying to create an expression for the magnetic force (from the magnetic field of the coil due to the induced current) as a function of velocity.

LeafNinja
I am not sure what simplifying assumptions could be made to calculate the force of eddy currents, but the magnetic field produced by the eddy currents should be small compared to the magnetic field produced by current through the coil. But good luck!