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Assume a conducting loop in a uniform magnetic field \(\vec B\).
Remember: when we assign a flat surface and choose \(d\vec A\) upwards, the magnetic flux is defined as
$$ \phi_B = \iint_R \vec B \cdot d\vec A \nonumber $$
The surface element \(dA = dx\,dy\), and \(d\vec A=\hat n\,dx\,dy\). The dot-product \(\vec B\cdot\hat n\) equals \(|\vec B|.1.\cos\theta\) $$ \begin{align} \phi_B &= \iint_R B\,\cos\theta\,dx\,dy \nonumber \\ &= B\,\cos\theta\int_0^y\int_0^x dx\,dy \nonumber \\ &= xy\,B\,\cos\theta \label{eq:phib} \end{align} \nonumber $$
According to Faraday, the time derivative of \(\phi_B\) determines the EMF (\(\mathcal E\)).
If the flux is positive and is getting larger positive, then based on Lenz’s law the current will be clockwise. This current will create an induced magnetic field \(\vec B_{ind}\) that is down.
Changing \(\text{ }d\phi_B/dt\)
We can change \(d\phi_B/dt\) by changing
- the magnetic field strength, (\(B\))
- the angle (\(\theta\))
- the area (\(xy\))
While looking at Lenz’s law, we already examined a change in magnetic field strength. Here we will examine a change in angle, and a change in area.
Changing the angle\(\text{ }\theta\)
To change the angle \(\theta\), we will rotate the loop around the \(y\)-axis with angular frequency \(\omega = 2\pi/\text{period}\)
The angle \(\theta\) is a function of \(\omega t\). We choose \(\theta_0\) so that at \(t=0\) the angle \(\theta=0\) $$ \theta = \bcancel{\theta_0} + \omega t \nonumber $$
Substituting this \(\theta\) in equation \(\eqref{eq:phib}\), and assigning the area \(A=xy\) $$ \begin{align*} \phi_B &= xy\,B\,\cos \omega t \\ &= A\,B\,\cos \omega t \end{align*} $$
The EMF is minus the derivative of \(\phi_B\) $$ \begin{align*} \mathcal E &= -\frac{d\phi_B}{dt} \\ \Rightarrow \mathcal E(t) &= -A\,B\,\frac{d}{dt}\cos \omega t \end{align*} $$
The EMF as a result of rotating the loop follows as $$ \shaded{ \mathcal E(t) = A\,B\,\omega\,\sin\omega t } \tag{EMF rotating} \label{eq:emfrotating} $$
Note that both the amplitute and period depend on the angular frequency. As the loop rotates, the current is going to alternate in a sinusoidal fashion. We call this alternating current (AC).
The induced current in the wire with resistance \(R\) is a result of the EMF \begin{align*} i(t) &= \frac{\mathcal E(t)}{R} \\ &= \frac{AB}{R}\omega\,\sin\omega t \end{align*}
This is the idea behind an alternating current generator. It has a permanent magnet, and you rotate conductor loops (windings), through the magnetic field. The higher the magnetic field, the faster you rotate, the more windings, bigger area of the loop .. the higher the EMF.
Multiple windings
If it was a closed loop with two windings, we would assign a surface with two layers. One is lower, and the other comes out at top. So, the magnetic flux will double, because the magnetic field penetrates the surface twice. You get twice the EMF.
On other words: if you have \(N\) windings in one closed loop, then the EMF would be \(N\) times larger.
Changing the area \(\text{ }xy\)
We can also change the area using a crossbar. Assuming a uniform magnetic field \(\vec B\)
We find the magnetic flux using on equation \(\eqref{eq:phib}\). We are varying the width (\(l\)) while keeping the angle \(\theta\) is zero. $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \phi_B &= xl\,B\cos\theta, & \left(\vec B\parallelsum d\vec A\right) \\ &= xl\,B \end{align*} $$
With \(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec B\parallelsum d\vec A\) and pointing in the same direction, the flux will be positive. Assuming the crossbar moves to the right with velocity \(v\), the flux will increase. According to Lenz’s law, the current will be clockwise
The EMF is minus the derivative of \(\phi_B\) $$ \begin{align*} \mathcal E &= -\frac{d\phi_B}{dt} \\ \Rightarrow \mathcal E(t) &= -lB\frac{dx}{dt}, & \left(\tfrac{dx}{dt} = v\right) \\ &= lBv \end{align*} $$
The EMF as a result of moving the crossbar follows as $$ \shaded{ \mathcal E(t) = l\,B\,v } \tag{EMF crossbar} $$
EMF without using Faraday’s law
We can alse determine the EMF without using Faraday’s law.
Looking at the left rod, the current is coming out of the screen, with the magnetic field pointing up. So, the Lorenz force is in the direction of \(\vec I\times \vec B\) (to the left)
With \(\vec I \perp \vec B\), the magnitude of the Lorenz force \(F_L\) is $$ F_L = I\,l\,B \nonumber $$
This is force that we have to apply to pull the crossbar to the right $$ F_{wl} = I\,l\,B \label{eq:fwl} $$
With \(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} F_{wl} \parallelsum \vec v\), we have to do work to move the crossbar to the right. This work comes out in the form of heat, in the resistance of this conductor. Because we’re creating an EMF, a current will flow and the dissipate the power $$ P = \mathcal E\,I = I^2 R \nonumber $$
If we change the direction by pushing the crossbar to the left, the current is going to change direction. So the Lorenz force will also flip over, and the force for us will flip over. Again we have to do positive work $$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} W &= \vec F\cdot d\vec x, & \left(P = \tfrac{dW}{dt}\right) \\ \Rightarrow P &= \vec F\cdot \frac{d\vec x}{dt}, & \left(\tfrac{d\vec x}{dt} = \vec v\right) \\ &= \vec F\cdot \vec v, & \left(\vec F\parallelsum\vec v\right) \\ &= F\,v \end{align*} $$
Substituting the force from equation \(\eqref{eq:fwl}\), the power that we generate is $$ \begin{align*} P &= I\,l\,B\,v, & \left(P = \mathcal E I\right) \\ \mathcal E\,\bcancel{I} &= \bcancel{I}\,l\,B\,v \end{align*} $$
This EMF is exactly what we found before $$ \shaded{ \mathcal E(t) = l\,B\,v } \tag{EMF crossbar} $$
Eddy currents
If we move a conducting disk into a magnetic field, then in the part of the disk moving under the magnet, the magnetic field will increase
This creates a clockwise electric field in the disk. This field induces a clockwise flow of electric current. So it produces an magnetic field in the opposite direction to oppose the change in magnetic flux
These currents are called eddy currents, after the localized water eddies in fluid dynamics. These currents produces heat $$ P = \mathcal E\,I = I^2 R \nonumber $$
The Lorenz force is in the direction of \(\vec I\times \vec B\), to the left. It opposes the motion. As a consequence, the disk will slow down, at the expense of kinetic energy. Heat is produced and it won’t go as fast through this field as then there was no magnetic field. We call that magnetic braking.
3-phase Motors
Generator
As we rotate about the axis, we’re going to get an induced EMF. We derived the induced EMF as equation \(\eqref{eq:emfrotating}\) $$ \mathcal E(t) = xy\,B\,\omega\,\sin\omega t \nonumber $$
Now we are going to add two more loops, which are not electrically connected to each other.
If you look from the front, you will see the first loop (1). The second loop (2) that is rotated 120°, and the third loop (3) again rotated 120°
Each one of those will give a sinusoidal EMF, but they are offset in phase by 120 degrees. We call this three phase current
If the period of (1) is \(60\,\rm{Hz}\), then the period of (2) and (3) is also \(60\,\rm{Hz}\).
Synchronous motor
Suppose, you’re looking down onto a horizontal table, and we have one solenoid (1) that we feed the current (1) going clockwise. Then we have another one (2), which is rotated 120°. We are going to feed that current (2). The same for solenoid (3), that we feed current (3).
The three phase current will produce a rotating magnetic field.
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At the moment the current through number (1) reaches a maximum, the current in (2) and (3) is two times smaller. But the vectorial sum of the magnetic field produced by (2) and (3) is in the same direction, so the net magnetic field is in the same direction.
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One third of a period later in time: now the current in (2) reaches a maximum, and the vectorial sum of (1) and (3) is in the same direction as well.
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Again one third of a period later: now the current in (3) reaches a maximum, and the vectorial sum of (1) and (2) is in the same direction as well.
We have created is now a rotating magnetic field. In one period, it rotates 360°. It rotates around in the period of the alternating current. If we put a magnet in the middle, then this magnet will to follow the magnetic field. We call this a synchronous motor.
So, the rotor of such a synchronous motor is magnet, and it rotates with the frequency of the alternating current.
Induction motor
For the rotor, we can also use a conducting sphere. The rotating magnetic field causes a continuous magnetic flux change with the surface of the sphere, and it’s going to run eddy currents. The eddy currents going around will give rise to a magnetic field, that causes a torque on the current. Similar to when we discussed earlier the idea of a synchronous motor.
So, it starts to torque up the sphere. You’re going to get a torque which is always be in the same direction, and the sphere object will start to rotate. We call this an induction motor
The speed of the induction motor depends on the conducting object. If it is a sphere, it will probably come very close to frequency the current, because a sphere has many possibilities for eddy currents to run around. Whereas if you take a ring, and you try to spin that in a rotating magnetic field, then the paths that are available for eddy currents are very limited. They can only go around in a ring.
Many of the stationary tools that you find in people’s workshops are induction motors. A table saw, drill press and electric grass mowers are induction motors.
2-phase motor
We can also make an induction motor that runs on 2-phase AC current. This is done using two solenoids, 90° apart.
The magnetic field will rotate 360° during one period of the AC current
- When the magnetic field is maximum in (1), it is zero in (2).
- When it is maximum in (2), it is zero in (1).
- When it is minimum in (1), it is zero in (2)
- When it is minimum in (2), it is zero in (1).
If we put a conductor in the middle, we’re going to have eddy currents, because we have change of magnetic flux all the time. These eddy currents will cause a torque on the conductor. The torque will always be in the same direction and the conductor is going to rotate.
