# Relationship between B and induced EMF using Faraday's Law

• Einstein44
Actually I had in my mind something like an AC generator where we control the angular velocity of rotation, I didn't read the OP carefully enough.

#### Einstein44

Homework Statement
What is the relationship between increasing the magnetic field strength and the induced emf as a magnet falls through a copper coil? Does anyone know it? I believe this would be increasing in a linear manner, but I wanted to be sure about that.
thanks
Relevant Equations
Faradays Law equation in differential form has to be used in this case.
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It is easier to see what's going on if you use Faraday's law in integral form. The instantaneous flux through the coil is a function of the distance to the magnet ##z## and the radius of the coil ##a##, i.e. ##\Phi(z,a).## You get that function by integrating the normal component of the magnetic field over the area of the loop ##S##, $$\Phi(z,a)= \int_S \vec B\cdot \hat n~dS$$Then you use Faraday's law to say $$\text{emf}=-\frac{d \Phi(z,a)}{dt}=-\frac{d \Phi(z,a)}{dz}\frac{dz}{dt}=-\frac{d \Phi(z,a)}{dz}v_z.$$I am not sure what you mean by "linear manner", but the expression is linear in ##v_z.## Getting the function ##\Phi(z,a)## is another issue. In general, elliptic integrals are involved that cannot be cast in closed form. You need to make some simplifying assumptions to avoid them.

Delta2 and Einstein44
Somehow I feel the OP asks:"What happens if I double the strength of B?" "Do I get double emf?"

Einstein44
Gordianus said:
Somehow I feel the OP asks:"What happens if I double the strength of B?" "Do I get double emf?"
Yes exactly

kuruman said:
It is easier to see what's going on if you use Faraday's law in integral form. The instantaneous flux through the coil is a function of the distance to the magnet ##z## and the radius of the coil ##a##, i.e. ##\Phi(z,a).## You get that function by integrating the normal component of the magnetic field over the area of the loop ##S##, $$\Phi(z,a)= \int_S \vec B\cdot \hat n~dS$$Then you use Faraday's law to say $$\text{emf}=-\frac{d \Phi(z,a)}{dt}=-\frac{d \Phi(z,a)}{dz}\frac{dz}{dt}=-\frac{d \Phi(z,a)}{dz}v_z.$$I am not sure what you mean by "linear manner", but the expression is linear in ##v_z.## Getting the function ##\Phi(z,a)## is another issue. In general, elliptic integrals are involved that cannot be cast in closed form. You need to make some simplifying assumptions to avoid them.
I was more trying to predict how the graph of B vs emf induced would look like and make the prediction using reasoning from the equations. I am not exactly sure what you did there.

Suppose the field around the magnet is given by ##\vec{B}(\vec{r})## and gives a flux $$\Phi=\iint_A \vec{B}\cdot d\vec{A}$$. If you replace the magnet by a stronger magnet such that the field around the new manget is ##\vec{B'}=k\vec{B}## where ##k>1## then it is easy to prove that the new flux $$\Phi'=k\Phi$$ and hence the new emf $$\mathcal{E}'=\frac{d\Phi'}{dt}=k\frac{d\Phi}{dt}=k\mathcal{E}$$, so yes the relationship is linear.

Einstein44
Einstein44 said:
I was more trying to predict how the graph of B vs emf induced would look like and make the prediction using reasoning from the equations. I am not exactly sure what you did there.

Delta2 said:
Suppose the field around the magnet is given by ##\vec{B}(\vec{r})## and gives a flux $$\Phi=\iint_A \vec{B}\cdot d\vec{A}$$. If you replace the magnet by a stronger magnet such that the field around the new manget is ##\vec{B'}=k\vec{B}## where ##k>1## then it is easy to prove that the new flux $$\Phi'=k\Phi$$ and hence the new emf $$\mathcal{E}'=\frac{d\Phi'}{dt}=k\frac{d\Phi}{dt}=k\mathcal{E}$$, so yes the relationship is linear.
Let me disagree a bit. Even though the flux increases linearly with B, the speed of the falling magnet isn't the same. The stronger the magnet, the lower the terminal speed.

Einstein44
Gordianus said:
Let me disagree a bit. Even though the flux increases linearly with B, the speed of the falling magnet isn't the same. The stronger the magnet, the lower the terminal speed.
I am afraid that is correct. I assumed that the speed of the magnet is controlled by other means and that it follows the same profile in both cases.

Actually I had in my mind something like an AC generator where we control the angular velocity of rotation, I didn't read the OP carefully enough.

Last edited:
Gordianus said:
Let me disagree a bit. Even though the flux increases linearly with B, the speed of the falling magnet isn't the same. The stronger the magnet, the lower the terminal speed.
Nobody said anything about terminal velocity. Usually, when one considers what changes might occur when one parameter is varied it is assumed that "all else remains the same". If the field is doubled, the emf is doubled all else remaining the same, and that includes the instantaneous velocity.

Einstein44 and Delta2
Delta2 said:
Suppose the field around the magnet is given by ##\vec{B}(\vec{r})## and gives a flux $$\Phi=\iint_A \vec{B}\cdot d\vec{A}$$. If you replace the magnet by a stronger magnet such that the field around the new manget is ##\vec{B'}=k\vec{B}## where ##k>1## then it is easy to prove that the new flux $$\Phi'=k\Phi$$ and hence the new emf $$\mathcal{E}'=\frac{d\Phi'}{dt}=k\frac{d\Phi}{dt}=k\mathcal{E}$$, so yes the relationship is linear.
Thank you very much, this should answer my question :)

Delta2