# Electromagnetic Induction

Electromagnetic induction states that a change in the magnetic flux \( \class{purple}{B} \) in a closed electrical circuit induces a voltage \( U_{\mathrm{ind}} \) or a current \( I_{\mathrm{ind}} \) in this circuit: $$U_{\text{ind}} ~=~ - A \, \frac{\class{violet}{\Delta B}}{\Delta t}$$

- Here, \( U_{\text{ind}} \) is the induced voltage. This electrical voltage is formed, for example, between the end points of a conductor loop when the magnetic field \( B \) penetrating the conductor loop is changed.
- Magnetic flux density \( B \) enclosed by the conductor loop, which is changed by the value \( \Delta B \). If this flux density \( B \) changes over time, i.e. \( \Delta B \neq 0 \), then an induction voltage or induction current arises in the conductor loop.
- Time period \( \Delta t \) within which the magnetic flux density has changed by the value \( \Delta B \).
- Area \( A \) enclosed by a conductor loop, for example. According to this formula, \( A \) is not changed, i.e. in this case it is assumed that the area remains unchanged.

The induced voltage \( U_{\mathrm{ind}} \) can also arise due to a change in area \( \Delta A \). Here we assume that the magnetic field does not change. Only the area that is penetrated by the magnetic field lines changes: $$U_{\text{ind}} ~=~ - \class{violet}{B} \, \frac{A}{\Delta t}$$

In the case of a coil with \(N\) turns, the induced voltage generated is \(N\)-times as high: $$U_{\text{ind}} ~=~ - N \, \class{violet}{B} \, \frac{A}{\Delta t}$$

## How does a dynamo work?

With a dynamo (e.g. a bicycle dynamo) you convert mechanical work (here: pedaling) into electrical energy.

How does it work? A dynamo consists of a coil with a certain number of turns \( N \) and a cross-sectional area \( A \). This coil is rotated in the magnetic field \( B \) of a permanent magnet, which generates an alternating voltage \( U_{\text{ind}}(t) \) according to Faraday's law of induction 1

.

If the coil rotates at a fixed angular frequency \( \omega \), the magnetic flux through the coil changes as follows:

All you have to do is to differentiate the magnetic flux 2

with respect to time and you get the voltage induced by the dynamo:

You can then use this induction voltage to light up a small lamp, for example.

## Exercises with Solutions

Use this formula eBook if you have problems with physics problems.### Exercise: A Conductor Loop Falling into a Magnetic Field

A very very long, rectangular conductor loop is falling *vertically* into a *constant* **magnetic field** \( B ~=~ 2 \, \text{T} \), which is directed into the plane. The conductor loop has a **resistance** \( R ~=~ 20 \, \Omega \) and a **width** \( w ~=~ 10 \, \text{cm} \).

As the conductor loop is falling all the time, it is being accelerated. At a certain time \( t_1 \), the conductor loop has reached a **velocity** of \( v(t_1) ~=~ 2 \, \frac{\text m}{\text s} \).

- Due to the change in magnetic flux through the area of the coil, an
**induction current**\( I \) is induced in the conductor loop. In which direction does the conventional current flow? - What is the magnitude of the current \( I(t_1) \) at time \( t_1 \)?
- Does the conductor loop experience an acceleration equal to the gravitational acceleration \( g \)?

#### Solution to Exercise #1

The magnetic field \( B \) is directed into the plane. Thus, the magnetic flux through the area \( A \) of the conductor loop increases into the plane. According to Lenz's law, nature tries to oppose this increase in magnetic flux. Therefore, an induction current \( I \) is induced in the conductor loop. The direction of the current must be such that the magnetic field \( B_{\text{ind}} \) produced by the induction current tries to prevent the increasing magnetic flux. Hence, the induced magnetic field \( B_{\text{ind}} \) must point out of the plane.

Apply the right-hand corkscrew rule. Curl your fingers along the conductor loop so that your thumb points out of the plane. This is the direction of the induced magnetic field. Consequently, the curled fingers point counterclockwise. Since these fingers indicate the direction of the current, the induction current must flow counterclockwise.

#### Solution to Exercise #2

To determine the induced current \( I \) at time \( t_1 \), we use Faraday's law:
`1
$$ U(t) ~=~ - \frac{ \text{d} \Phi}{ \text{d}t } $$
`

This gives us the relationship between the induction voltage \( U(t) \) and the *rate of change* of the magnetic flux \( \Phi(t) \), represented by the time derivative.

Since we are looking for the induction current \( I \), we use Ohm's law \( U ~=~ R \, I \) to express Faraday's law in terms of current. To do this, we substitute Ohm's law for voltage:
`2
$$ R \, I(t) ~=~ - \frac{ \text{d} \Phi}{ \text{d}t } $$
`
and move \( R \) to the right-hand side:
`3
$$ I(t) ~=~ - \frac{1}{R} \, \frac{ \text{d} \Phi}{ \text{d}t } $$
`

Now, we need to determine how exactly the magnetic flux \( \Phi \) changes over time. We can express the magnetic flux as \( \Phi(t) = B \, A(t) \). The magnetic field \( B \) into which the conductor loop falls is constant according to the task, hence time-independent.

However, the area \( A(t) \) through which the magnetic flux penetrates within the conductor loop increases over time because the loop is falling, allowing more magnetic flux to penetrate its interior.

Therefore, we can pull the constant magnetic field out of the time derivative and only consider the time derivative of the magnetically penetrated area \( A(t) \) (see Illustration 2):
`4
$$ I(t) ~=~ - \frac{B}{R} \, \frac{ \text{d} A}{ \text{d}t } $$
`

The area is not explicitly given, but we can break it down into known quantities. We know that the loop has a width \( w \). Additionally, we denote the current height of the penetrated area as \( h(t) \). The width \( w \) of the loop remains constant during falling. Therefore, it is time-independent. However, the current height \( h(t) \) of the area increases over time because the loop is falling, causing the penetrated area in the vertical height to increase. The product of the width \( w \) and the height \( h(t) \) yields the rectangular area \( A(t) \). We can substitute this into equation 4

, pulling the constant width \( w \) out of the derivative:
`5
$$ I(t) ~=~ - \frac{B \, w}{R} \, \frac{ \text{d} h(t)}{ \text{d}t } $$
`

The velocity \( v(t) \) is generally defined as the time derivative of displacement. This is exactly what we have in equation 5

. There, the derivative of the height (i.e., displacement) with respect to time \( t \) is given. Therefore, we can replace the time derivative with the velocity \( v(t) \):
`6
$$ I(t) ~=~ - \frac{B \, w}{R} \, v(t) $$
`

With this formula, we can calculate the current at any given time if we know the velocity \( v(t) \) at the same time. In the task, we know the velocity \( v(t_1) = 2 \, \frac{\text m}{\text s} \) at time \( t_1 \). Thus, we can calculate the current \( I(t_1) \) at this time \( t_1 \):

`7 $$ I(t_1) ~=~ - \frac{B \, w}{R} \, v(t_1) $$`

There are no unknowns left. Now, we just need to substitute concrete values to calculate the current:
`8
\begin{align}
I(t_1) &~=~ - \frac{2 \, \text{T} ~\cdot~ 0.1 \, \text{m}}{ 20 \, \Omega } ~\cdot~ 2 \, \frac{\text m}{\text s} \\\\
&~=~ - 0.02 \, \text{A}
\end{align}
`

#### Solution to Exercise #3

When falling, the conductor loop experiences the following gravitational force downward toward the Earth:
`9
$$ F_{\text g} ~=~ m \, g $$
`

Since there is a current flowing in the lower horizontal bar of the conductor loop and it is situated in an external magnetic field \( B \), a Lorentz force acts on this portion of the conductor. When applying the right-hand rule:

- Thumb points in the direction of the current
- Index finger points into the plane