Distance from one period to the next, ##e## in helical path.

[Potatochip911]

Homework Statement

A source injects an electron of speed $v$ into a uniform magnetic field of magnitude $B$. The velocity of the electron makes an angle $\theta$ with the direction of the magnetic field. Find the distance $d$ from the point of injection at which the electron next crosses the field that passes through the injection point.

Homework Equations

$F_B=m\frac{v^2}{r}$
$T=\frac{2\pi r}{v}$

The Attempt at a Solution

We want to find how much vertical displacement there is in one period essentially. To find $T$ in terms on known quantities:

$$
F_B=q(v\times B)=qvB\sin\theta=\frac{v^2}{r}\Longrightarrow r=\frac{mv}{qB\sin\theta} \\
T=\frac{2\pi r}{v}=\frac{2\pi m}{qB\sin\theta}
$$

Now the vertical displacement would be given by $v\cdot\cos\theta$ multiplied by the time (period in this case) Therefore:
$$
d=v\cos\theta\cdot \frac{2\pi m}{qB\sin\theta}=\frac{2\pi m v \mbox{ cotan }\theta}{qB}
$$

In my solution manual however they have completely neglected $F_B=q(v\times B)$ and just used the result for when $\theta=90$ ($F_B=qvB$), in the textbook the angle given is not equal to 90 so this isn't the case. Is there something I'm missing here or did they solve it incorrectly?

Edit: Some errors in my derivation:

$$F_B=qvB\sin\theta=m\frac{v^2}{r}$$

the velocity $v^2/r$ in this term is the velocity that is perpendicular to the force so it becomes: $qvB\sin\theta=m\frac{(v\sin\theta)^2}{r}$ This solves the problem.

 

[gneill]

Closed: resolved by OP.