Question

3- A wire forms a coil (helix) of radius 5 [cm] and rises a height of n [cm] for each full turn. One end of the wire is a point (5,0,0) and the other end is at (0,5, An electric field is given by: 131T Calculate the total emf produced by this field along the helix path. Ill

Answer 1

It would be convenient to work in cylindrical coordinates. General line element is

$\vec{dl}=dr\hat{r}+rd\theta\hat{\theta}+dz\hat{z}$

The electric field becomes

$\vec{E}=-x\hat{x}-z\hat{y}-xy\hat{z}=-R\cos\theta(\hat{r}\cos\theta-\hat{\theta}\sin\theta)-z(\hat{r}\sin\theta+\hat{\theta}\cos\theta)-R^2\sin\theta\cos\theta\hat{z}$

$\vec{E}\cdot\vec{dl}=(r\cos^2\theta+z\sin\theta)dr+(r\sin\theta\cos\theta-z\cos\theta)rd\theta-r^2\sin\theta\cos\theta dz$

Now we defined the curve.

$r=5, \frac{\mathrm{d} z}{\mathrm{d} \theta}=\frac{\pi}{2\pi}=\frac{1}{2}\Rightarrow dz=\frac{d\theta}{2}\Rightarrow z=\frac{\theta}{2}$

$\int\vec{E}\cdot\vec{dl}=\int(5\sin\theta\cos\theta-\frac{1}{2}\theta\cos\theta)5d\theta-\frac{25}{2}\sin\theta\cos\theta d\theta$

$\int\vec{E}\cdot\vec{dl}=\int_{0}^{13\pi/2}\left (\frac{25}{2}\sin\theta\cos\theta-\frac{5}{2}\theta\cos\theta \right )d\theta$

$\frac{35}{4}-\frac{65\pi}{4}$