Emf in the Archimedean spiral

Because the number of turns is big, we suppose each turn is nearly a circle.

The emf induced in the circuit is: $E=- \frac{d\Phi}{dt}=- \frac{dB}{dt} \Sigma S$ , where $\Sigma S$ is the total area of all circles.

Because the number of circles is big, we can calculate $\Sigma S$ by calculate the average area of all circles: $S_{average}= \int\limits_0^R \frac{\pi r^2 dr}{R}=\frac{\pi R^2}{3}$ .

Therefore, $\Sigma S=N S_{average}=\frac{N \pi R^2}{3}$ .

So $E= \omega B_0 \sin(\omega t) \frac{N \pi R^2}{3}$ .

The amplitude of E is $\frac{1}{3} \pi R^2 N \omega B_0=2.094 V$ .

Firstly, let's calculate the region's area $A$ :

In polar coordinates: $A = \int \frac{r^2d\theta}{2} = \int^{100*2\pi}_{0}\frac{b^2\theta^2d\theta}{2} = \frac{b^2(200\pi)^3}{6}$ and since that $r = 10cm$ when $\theta = 200\pi$ , then $b=\frac{0.1}{200\pi}$

So, the area is $A = \frac{(0.1)^2*200\pi}{6} = 1.0471975512 m^2$

Now, let's calculate the magnetic flux: $\phi = B(t)A$

Using the Faraday's Law, we have the emf induced : $\epsilon = \frac{d\phi}{dt}$

So, $\epsilon = -B_{o}\omega Asin(\omega t)$ . Therefore, the amplitude of the emf in Volts induced is $\epsilon_{max} = B_{o}\omega A= 1*10^{-6}*2*10^{6}*1.0471975512 = 2.094 V$

The induced emf by changing magnetic field is given by

$\varepsilon = -N \frac{d(\Phi)}{dt}$ where $\Phi = BA_{\perp}.$

The area of the spiral is one-third of the circle enclosing it so that

$A_\perp = \frac{\pi}{3}r^2 = \frac{\pi}{3}(0.1 m)^2 = \frac{\pi}{300} m^2.$

Solving for $\varepsilon$ ,

$\varepsilon = -N \frac{d(\Phi)}{dt} = -N \frac{d(BA_\perp)}{dt} = -NB_0A_\perp \frac{d(\cos(\omega t))}{dt}$

$\varepsilon = NB_0A_\perp\omega\sin(\omega t).$

Substituting the given values, the amplitude of the induced emf is

$\varepsilon_m = (100)(1\times10^{-6} \mu T)(\frac{\pi}{300} m^2)(2\times10^6 s^{-1}) = 2.094$ $volts.$

From Faraday's Law, we have $$\mathcal{E}=-\frac{d\Phi}{dt}=-\frac{d}{dt}\left[BA\right].$$ We are given $$B=B 0 \cos{\left(\omega t\right)},$$ and we can calculate the total are inside the circuit by evaluating a polar integral of $r=\frac{R}{2\pi N}\theta$ , which is the general form for an Archimedean spiral with outer radius $R$ and $N$ turns. $$A=\frac{1}{2}\int {0}^{2\pi N}\left({\frac{R}{2\pi N}\theta}\right)^2 \ d\theta=\frac{\pi}{3} N R^2$$ Substituting back into Faraday's Law gives $$\mathcal{E}=-\frac{d}{dt}\left[\frac{\pi}{3} N R^2 B_0 \cos{\left({\omega t}\right)}\right]$$

$$=\frac{\pi}{3} N R^2 B_0 \omega \sin{\left({\omega t}\right)}.$$

The amplitude $a$ of this function is just the coefficient of the sine function.

$$a=\frac{\pi}{3} N R^2 B_0 \omega $$ $$=\frac{\pi}{3} \left(100\right) \left(0.10 \ \mathrm{m}\right)^2 \left(1 \times 10^{-6} \ \mathrm{T}\right) \left(2 \times 10^6 \mathrm{s^{-1}}\right) $$ $$= \frac{2\pi}{3} \ \mathrm{V} \approx \fbox{2.094 V}. $$

First we determine the $b$ in the equation $r =b\theta$ . We note that because there is 100 spirals, that means that $\theta = 100(2\pi)$ when $r = 0.1m$ . Thus we get the equation as $r = \frac{1}{2000\pi} \theta \ .$ The flux through the surface is given as \Phi_B = \iint\limits_A \bf{B}\, \dot\, \bf{dA} \ \ . We know that the magnetic field is parallel to the normal of the surface, and that $B$ has no dependence on the area, so we can write the equation as $\Phi_B = B \iint\limits_A dA$ Doing the integral in polar coordinates, we have $\Phi_B = B \int_0^{200\pi}\!\!\!\! \int_0^{b\theta}\!\! rdrd\theta \ .$ Solving this, we end up with $\Phi_B = \frac{\pi}{3}B \ .$ Now to find the amplitude of the emf, we use Faraday's law of induction, $\mathcal{E} = -\frac{d\Phi_B}{dt} \ .$ Since $B = B_0cos\omega t$ , we obtain $\mathcal{E} = \frac{\pi}{3}B_0\omega sin\omega t \ .$ The amplitude of the emf would thus be $\frac{\pi}{3}B_0\omega = 2.09\, V \ .$

Let's find $b$ . If $N = 100$ we'll have $\theta = 2\pi\cdot 100$ . So: $$ r = b\theta \Rightarrow R = b\cdot 200\pi \Rightarrow b = \frac{1}{20\pi} $$ Calculating $\epsilon$ : $$ \phi = B\cdot A \Rightarrow \epsilon = -\frac{d\phi}{dt} = - A\cdot\frac{dB}{dt} $$ $$ \epsilon = AB 0\omega\sin(\omega t) $$ So, we observe that the amplitude of emf is $AB_0\omega$ . Here we need to calculate the area of the spiral: $$ A = \int\int rdrd\theta = \int {\theta = 0}^{200\pi}\frac{r^2}{2}d\theta = \int {\theta = 0}^{200\pi}\frac{b^2\theta^2}{2}d\theta = \frac{b^2(200\pi)^3}{6} $$ Finally, applying the corresponding values: $$ \epsilon {max} = AB_0\omega = 2.094 $$

By Faraday's Law of Electromagnetic Induction, $\varepsilon = -N \frac {d \Phi_B} {dt}$ . Since $\Phi_B=BA$ and the area enclosed by an Archimedean spiral is $\frac{1}{3}$ the area of the circle enclosing it, we have that $\varepsilon = -N \frac {d (B_0 cos(\omega t)\frac{1}{3} \pi r^2)} {dt}$ . Therefore, we have that $\varepsilon = N B_0 \frac{1}{3} \pi r^2 \omega sin(\omega t)$ , so the amplitude of the induced emf is $N B_0 \frac{1}{3} \pi r^2 \omega$ . Substituting in our given values, we get that our answer is $100(10^{-6})(\frac{1}{3})(\pi)(0.1)^2(2*10^{6}) \approx \fbox{2.094}$

According to Maxwell's equations, we can calculate the EMF as $\mathcal{E}=\frac{\mathrm{d}}{\mathrm{d}t}\oint{\mathbf{B}\,\mathrm{d}\mathbf{S}}$ . As the induction of the magnetic field is constant over space (i.e. $\nabla\mathbf{B}=0$ ), we can take it outside of the integral. Also note, that since the spiral is flat, $\mathrm{d}\mathbf{S}$ is always parallel to $\mathbf{B}$ , allowing us to substitute $B\,\mathrm{d}S$ for $\mathbf{B}\,\mathrm{d}\mathbf{S}$ .

The area enclosed by the spiral can be calculated as $S=\frac{1}{2}\int_0^{2\pi\times 99.75}{b^2 \vartheta^2\,\mathrm{d}\vartheta}=\frac{b^2}{2}\frac{(2\pi\times 99.75)^3}{3}$ , where we took into account that actually (according to the figure) the spiral makes a quarter of a turn less than stated.

$b$ can be calculated from the equation for a spiral, substituting $r=0.1\,\mathrm{m}$ and $\vartheta=2\pi\times99.75$ . So, $b=\frac{1}{1995\pi}$ . Combining everything we have got thus far, $\mathcal{E}_{\mathrm{amp}}=B_0 S\omega=10^{-6}\times\frac{133\pi}{400}\times 2\times10^{6}=2.089\,(\mathrm{V})$ .

The Maxwell-Faraday relationship governs the behavior of yet another closed wire in a field!

$\displaystyle -\frac{d\phi_B}{dt} = \mathcal{E}$

Instead of a simple circle or square, we have the spiral of Archimedes which has a series of overlapping elements, each subsequent turn of the spiral containing the area of the turn which precedes it. The field is uniform, so the flux is given by

$\phi_B = A_{eff}\vec{B}\cdot\hat{k}$

where $A_{eff}$ is the effective area of the entire spiral.

We can calculate the total effective area by integrating the radius around the $N$ turns. This is given by

$\displaystyle A_{eff} = \int\limits_{0}^{b\theta}\int\limits_{0}^{200\pi}r dr d\theta = \frac43 b^2N^3\pi^3$

We know that $r = b\theta$ and we also know that the radius at 100 turns ( $\displaystyle \theta = 200\pi$ ) is 0.1 cm , thus $\displaystyle b = \frac{1}{2000\pi}$ .

Plugging this all in, we find $\displaystyle A_{eff} = \frac{\pi}{3}$ .

Therefore, $\displaystyle \phi_B = \frac{\pi}{3}B_0\cos\omega t$ making the magnitude of $\displaystyle\frac{d\phi_B}{dt}$ equal to $\boxed{\displaystyle\frac{\pi}{3}B_0\omega = \frac{2\pi}{3} \approx 2.094}$ .