Week 14: Induction

Induction Graphs

In these notes, we will examine a few examples of changing magnetic fluxes and associated induced voltages. Recall from the previous notes that these are related by Faraday’s Law which says:

This is saying that the induced current is the negative slope of the magnetic flux. In other words, if the magnetic flux is increasing, then $V_{ind}$ will be negative, if the magnetic flux is decreasing, then $V_{ind}$ will be positive, and if the magnetic flux is constant, then $V_{ind} = 0$.

First let’s consider when an example where $\Phi_{B}$ rises and falls linearly with the same magnitude of slope:

From $t = 0$ to $t = 5$, $\Phi_{B}(t)$ has a constant positive slope, so $V_{ind}$ will be constant and negative. Conversely, from $t = 5$ to $t = 10$, $\Phi_{B}(t)$ has a constant negative slope, so $V_{ind}$ will be constant and positive.

Specifically, in this case $\Phi_{B}(t)$ is defined as:

$$ \Phi_{B}(t) = \left{ \begin{matrix} 2t & \text{if~}0 < t < 5 \

• 2t & \text{if~}5 < t < 10 \end{matrix} \right.\ $$

Which means $\frac{d\Phi_{B}}{dt}$ is:

$$ \frac{d\Phi_{B}}{dt} = \left{ \begin{matrix} 2 & \text{if~}0 < t < 5 \

• 2 & \text{if~}5 < t < 10 \end{matrix} \right.\ $$

Now we can multiply by $- 1$ because of the negative sign in Faraday’s law to find $V_{ind}$:

$$ V_{ind} = \left{ \begin{matrix}

• 2 & \text{if~}0 < t < 5 \ 2 & \text{if~}5 < t < 10 \end{matrix} \right.\ $$

Next, let’s consider an example with a few different slopes:

We can see that from $t = 0$ to $t = 10$, $\Phi_{B}(t)$ has a positive slope, so $V_{ind}$ is negative on that time interval. However, $\Phi_{B}(t)$ is steeper from $t = 5$ to $t = 10$, so $V_{ind}$ is more negative on that time interval than from $t = 0$ to $t = 5$. From $t = 10$ to $t = 15$, $\Phi_{B}(t)$ has a constant and negative slope, so $V_{ind}$ is constant and positive on that time interval. Specifically we have that:

$$ \Phi_{B}(t) = \left{ \begin{matrix} 2t & \text{if~}0 < t < 5 \ 5t - 15 & \text{if~}5 < t < 10 \

• 10t + 135 & \text{if~}10 < t < 15 \end{matrix} \right.\ $$

Which means $\frac{d\Phi_{B}}{dt}$ is:

$$ \frac{d\Phi_{B}}{dt} = \left{ \begin{matrix} 2 & \text{if~}0 < t < 5 \ 5 & \text{if~}5 < t < 10 \

• 10 & \text{if~}10 < t < 15 \end{matrix} \right.\ $$

Which finally means that $V_{ind}$ is:

$$ V_{ind} = \left{ \begin{matrix}

• 2 & \text{if~}0 < t < 5 \
• 5 & \text{if~}5 < t < 10 \ 10 & \text{if~}10 < t < 15 \end{matrix} \right.\ $$

Finally, let’s look at an example with a non-linear $\Phi_{B}(t)$:

$\Phi_{B}(t)$ looks like a quadratic centered about t = 2. We can see that while $\Phi_{B}(t)$ is decreasing ($0 < t < 2$), $V_{ind}$ is positive, and while $\Phi_{B}(t)$ is increasing ($2 < t < 8$), $V_{ind}$ is negative.

Specifically, in this case we have:

Taking a first derivative with respect to time yields:

Multiplying by $- 1$ to find $V_{ind}$ gives: