# EMF induced in solenoid by rotating magnet

I would like to have an equation relating the emf (or current) induced in a solenoid by a rotating magnet with angular velocity $$\omega$$ and the distance $$d$$ as shown in the figure.

I really don't know how complex this problem is, maybe some approximation/simplification is needed to get a result. Also, you can make any further assumption if needed (such as the amplitude $$B_0$$ of the magnetic induction in the center of the solenoid when the magnet and the solenoid are aligned, or the length of the solenoid, etc.)

Finally, I'm aware that the general equation is $$\epsilon = -\frac{d\phi_B}{dt}$$ but I can't see any way to get an expression for $$\phi_B$$ in order to compute the time-derivate.

• May I assume that the magnetic field strength is going to remain constant all the time at every point in the space in direction in which the magnetic field will be located at given point of time? Commented Jan 31, 2022 at 14:21
• @TejasDahake I'm not sure I understood your assumption. Which are the points you would assume to have constant $|B|$?
– Javi
Commented Jan 31, 2022 at 18:19
• I mean is it okay to take an assumption that the field lines originating from the magnet are parallel and equidistant from eachother that means the field strength is uniform in the direction in which the magnet is oriented at some instant because the magnet is rotating. Commented Jan 31, 2022 at 20:24
• @TejasDahake its ok, approximate solutions may be helpful too
– Javi
Commented Feb 1, 2022 at 14:04
• I've answerd your question you can check that out below. Commented Feb 3, 2022 at 9:57

I am taking some of the assumptions to keep the case ideal and simple to understand the first assumption I'm going to take is that the field lines originating from the magnet are parallel and equidistant from eachother that means the field strength is uniform in the direction in which the magnet is oriented at some instant because the magnet is rotating, secondly if the field strenght is constant (magnitude wise) then the distance from the magnet to the coil doesn't matter at all.

Since this is the case of rotating magnetic field then it's going to produce an Alternating Current no DC current let's understand how:-

As we know that

$$\phi_B = \vec{B} • \vec{A}$$

$$\phi_B = ABcos \theta$$

Since there are "$$N$$" turns in the solenoid the total Flux will also become "$$N$$" times so total Flux will come out to be:-

$$N \phi_B = NABcos \theta$$

Assigning $$N \phi_B$$ as $$\phi_0$$ So,

$$\phi_0 = NABcos \theta$$

By differentiating both side with time we will get

$$-\frac{d (\phi_0)}{dt} = \frac{d(NABcos \theta)}{dt}$$

Negative sign is because of:-

$$E_0 = -\frac{d \phi}{dt}$$

Since number of turns, magnetic field strength and area are constant all the time and due to the rotation of the magnetic field the angle between the field lines and the area vector is going to change so The final equation you are going to get is:-

$$E_0 = NBAsin \theta • \frac{d \theta}{dt}$$

Since $$\frac{d \theta}{dt}$$ is $$\omega$$ the equation will be

$$E_0 = NBA \omega sin \theta$$

Hence you got the equation for the induced EMF in the case of rotating magnet, and here when the the angle between the area vector and the field is 0 then no EMF will be induced, in the case if it's perpendicular then the induced EMF will be maximum and when the angle between them will be greater than $$180°$$ then the induced EMF will be negative or in other words if we can imagine a battery which is formed then it's orientation is going to change and you will get the current in the opposite direction as that of before.

And by dividing both the sides of the equation with the resistance of the wire then you can find the current as well, so there you go!

• Nice answer, so here $\theta = 0$ occurs when the magnet is parallel to the solenoid? And also $\theta = 180$ but with N and S switching sides. Am I right?
– Javi
Commented Feb 5, 2022 at 15:49
• You are right. Zero means the field vector and area vector are parallel to each other and 180° means they are anti-parallel or in other words they've switched the sides as you said! Commented Feb 5, 2022 at 18:16

For estimating the field strength of a bar magnet, you can replace it by a current carrying solenoid of the same size, shape, and dipole moment. To get the dipole moment of the magnet, you might suspend it on a thin thread, and let it oscillate as a torsion pendulum in the earths field. You can find the field at any point on the axis of a known current carring solenoid by integrating the contribution from each turn of current. This lets you estimate the maximum field at each point on the axis of the real solenoid when the bar magnet is lined up with the solenoid. If the real solenoid has a small radius, this field should from not change much as you move away from the axis, and you can use it to estimate the maximum flux through each turn of the solenoid. Adding these gives the total maximum flux. Finally, use this maximum flux as the amplitude of an oscillating flux from which you can find the amplitude of the induced emf.

• This is the case of bar magnet and as OP mentioned in his/her post some distance d seems like he/she is trying to tell us that the magnetic field is going to vary with distance so won't it affect the induced emf? Commented Jan 31, 2022 at 20:22
• The field strength and the flux through the solenoid will decrease with increasing distance, and this will effect the induced emf. Commented Feb 1, 2022 at 13:56
• @R.W.Bird this seems helpful. So I could get the magnetic dipole moment $m$ and then $m = n I A$ to get the current $I$ in the new solenoid that replaces the magnet. Is this what you mean? Furthermore, could you expand on how to get the dipole moment o perhaps point me to some source that explains it?
– Javi
Commented Feb 1, 2022 at 17:23
• A magnetic dipole in an external field is subject to a restoring torque which is proportional to the angular displacement. When combined with the rotational inertia, this determines the period of oscillation. Rather than trying to predict the emf from this system, I would suggest that you just build it and see how it behaves. Commented Feb 2, 2022 at 14:31

You need to make the following assumptions:

Distance between solenoid and magnet is small, thereby all field lines pass through the solenoid.

Thus when magnet's north is facing the solenoid, you get a flux of $$B×N×A$$. (B is magnetic field, N is number of turns in solenoid and A is area of each loop of solenoid)

When magnet's south is facing the solenoid, the flux becomes -$$B×N×A$$.

Change in flux = BNA - (-BNA) = 2BNA

Time required for half spin (so that north becomes south) will come out to be:

$$\frac{π}{\omega}$$

Thus induced EMF = change in flux / time required

$$EMF = 2BNA × \frac{\omega}{π}$$

• As I understand it, this is a constant expression that gives the mean voltage induced in the solenoid. Is this correct?
– Javi
Commented Feb 1, 2022 at 18:45
• You are right. This gives the mean voltage. However, with the available info, this seems like the answer one might be looking for. It also depends on the education level at which you are answering. For a high school student, this seems enough, for someone doing masters in physics, this answer will need refinement. :) Commented Feb 2, 2022 at 10:25
• yes that's true. Actually this question is about a real project (technical but not scientific) for which I have been asked for help. I believe this answer might be useful for this purpose. Thanks.
– Javi
Commented Feb 2, 2022 at 11:08
• @Whiskeyjack the assumption you took here that the distance must be very small actually doesn't matters at all because you haven't told anything about the field strenght if it is the function of distance or not, so most of the novices will interpret this thing as an arbitrary constant. Commented Feb 2, 2022 at 19:47