Electromagnetic Induction

The textbook states that the direction of induced current in each of the six cases of Fig. 6.15(a)–(f) is to be predicted from the figure itself using Lenz's law. Use Lenz's law: the induced current always opposes the change in magnetic flux producing it.

The applicable rule is Lenz's law: the induced current opposes the change in flux.

For a long solenoid, magnetic field inside is
$$B=\mu_0 n I$$
So the flux through the small loop is
$$\Phi = BA = \mu_0 n I A$$
Hence induced emf is
$$\varepsilon = A\mu_0 n \frac{dI}{dt}$$
Given:
- turns per cm = 15, so $n=15\times 100=1500\,\text{m}^{-1}$
- area $A=2.0\,\text{cm}^2=2.0\times10^{-4}\,\text{m}^2$
- current change $\Delta I=4.0-2.0=2.0\,\text{A}$
- time $\Delta t=0.1\,\text{s}$

So
$$\frac{dI}{dt}=\frac{2.0}{0.1}=20\,\text{A s}^{-1}$$
Now
$$\varepsilon = (2.0\times10^{-4})(4\pi\times10^{-7})(1500)(20)$$
$$=2.0\times10^{-4}\times 4\pi\times10^{-7}\times 3.0\times10^{4}$$
$$=2.0\times10^{-4}\times 3.77\times10^{-2} \approx 7.5\times10^{-6}\,\text{V}$$
So the induced emf is about $7.5\,\mu\text{V}$.

Use motional emf:
$$\varepsilon = Blv$$
Here
- $B=0.3\,\text{T}$
- velocity $v=1\,\text{cm s}^{-1}=0.01\,\text{m s}^{-1}$

The length $l$ is the side perpendicular to the motion.

### (a) Motion normal to the longer side
Then the side cutting the field lines is the longer side, $l=8\,\text{cm}=0.08\,\text{m}$.
$$\varepsilon = 0.3\times0.08\times0.01=2.4\times10^{-4}\,\text{V}$$
Time for the loop to leave the field region equals width in direction of motion divided by speed. If it moves normal to the longer side, it crosses the shorter side width $2\,\text{cm}=0.02\,\text{m}$:
$$t=\frac{0.02}{0.01}=2\,\text{s}$$

### (b) Motion normal to the shorter side
Then the effective length is the shorter side, $l=2\,\text{cm}=0.02\,\text{m}$.
$$\varepsilon = 0.3\times0.02\times0.01=6.0\times10^{-5}\,\text{V}$$
Now the loop crosses the longer side width $8\,\text{cm}=0.08\,\text{m}$:
$$t=\frac{0.08}{0.01}=8\,\text{s}$$
So the induced voltage is $2.4\times10^{-4}\,\text{V}$ for 2 s in (a), and $6.0\times10^{-5}\,\text{V}$ for 8 s in (b).