16.2 Faraday’s law of induction: lenz’s law  (Page 3/6)

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Making connections: conservation of energy

Lenz’s law is a manifestation of the conservation of energy. The induced emf produces a current that opposes the change in flux, because a change in flux means a change in energy. Energy can enter or leave, but not instantaneously. Lenz’s law is a consequence. As the change begins, the law says induction opposes and, thus, slows the change. In fact, if the induced emf were in the same direction as the change in flux, there would be a positive feedback that would give us free energy from no apparent source—conservation of energy would be violated.

Calculating emf: how great is the induced emf?

Calculate the magnitude of the induced emf when the magnet in [link] (a) is thrust into the coil, given the following information: the single loop coil has a radius of 6.00 cm and the average value of $B\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ (this is given, since the bar magnet’s field is complex) increases from 0.0500 T to 0.250 T in 0.100 s.

Strategy

To find the magnitude of emf, we use Faraday’s law of induction as stated by $\text{emf}=-N\frac{\Delta \Phi }{\Delta t}$ , but without the minus sign that indicates direction:

$\text{emf}=N\frac{\Delta \Phi }{\Delta t}\text{.}$

Solution

We are given that $N=1$ and $\Delta t=0\text{.}\text{100}\phantom{\rule{0.25em}{0ex}}\text{s}$ , but we must determine the change in flux $\Delta \Phi$ before we can find emf. Since the area of the loop is fixed, we see that

$\Delta \Phi =\Delta \left(\mathrm{BA}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)=A\Delta \left(B\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right).$

Now $\Delta \left(B\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)=0\text{.}\text{200 T}$ , since it was given that $B\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ changes from 0.0500 to 0.250 T. The area of the loop is $A={\mathrm{\pi r}}^{2}=\left(3\text{.}\text{14}\text{.}\text{.}\text{.}\right)\left(0\text{.}\text{060 m}{\right)}^{2}=1\text{.}\text{13}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ . Thus,

$\Delta \Phi =\left(\text{1.13}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)\left(0.200 T\right).$

Entering the determined values into the expression for emf gives

$\text{Emf}=N\frac{\Delta \Phi }{\Delta t}=\frac{\left(1.13×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)\left(0\text{.}\text{200}\phantom{\rule{0.25em}{0ex}}\text{T}\right)}{0\text{.}\text{100}\phantom{\rule{0.25em}{0ex}}\text{s}}=\text{22}\text{.}6\phantom{\rule{0.25em}{0ex}}\text{mV.}$

Discussion

While this is an easily measured voltage, it is certainly not large enough for most practical applications. More loops in the coil, a stronger magnet, and faster movement make induction the practical source of voltages that it is.

Play with a bar magnet and coils to learn about Faraday's law. Move a bar magnet near one or two coils to make a light bulb glow. View the magnetic field lines. A meter shows the direction and magnitude of the current. View the magnetic field lines or use a meter to show the direction and magnitude of the current. You can also play with electromagnets, generators and transformers!

Section summary

• Faraday’s law of induction states that the emf induced by a change in magnetic flux is
$\text{emf}=-N\frac{\Delta \Phi }{\Delta t}$

when flux changes by $\Delta \Phi$ in a time $\Delta t$ .

• If emf is induced in a coil, $N$ is its number of turns.
• The minus sign means that the emf creates a current $I$ and magnetic field $B$ that oppose the change in flux $\Delta \Phi$ —this opposition is known as Lenz’s law.

Conceptual questions

A person who works with large magnets sometimes places her head inside a strong field. She reports feeling dizzy as she quickly turns her head. How might this be associated with induction?

A particle accelerator sends high-velocity charged particles down an evacuated pipe. Explain how a coil of wire wrapped around the pipe could detect the passage of individual particles. Sketch a graph of the voltage output of the coil as a single particle passes through it.

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