Yes exactlyGordianus said:Somehow I feel the OP asks:"What happens if I double the strength of B?" "Do I get double emf?"
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.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.
Sorry, I misunderstood what you were asking. Your question has been adequately answered by @Delta2 in # 6 above.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.
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.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.
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.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.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.
Thank you very much, this should answer my question :)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.