EMF induced in solenoid by rotating magnet

Question

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.

Answer 1

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!

Answer 2

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.

Answer 3

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

$EMF = 2BNA × \frac{\omega}{π}$