Answer
- $r=5.25\times 10^{-5} \;\rm m$
- $p=3.3\times 10^{-4}\;\rm m$
Work Step by Step
Since the electron is moving on a circular path, so the force exerted on it is a centripetal force due to the magnetic field.
This force is given by
$$F_B=qB v\sin\theta =ma_r$$
Noting that $\theta=90^\circ$ since we need the perpendicular direction of the velocity of the magnetic field.
Thus,
$$ q B v_{\perp }=m\dfrac{v_{\perp}^2}{r}$$
Noting that the electron enters the magnetic field at an angle of $\theta=45^\circ$, so $v_{\perp}=v\sin45^\circ$
$$ q B =m\dfrac{v \sin45^\circ }{r}$$
where $q$ is the charge of the electron, and $m$ is its mass.
Solving for $r$;
$$ r= \dfrac{mv \sin45^\circ }{qB }$$
Plugging the known;
$$ r= \dfrac{ 9.11\times10^{-31}\cdot 3\times 10^6 \cdot \sin45^\circ }{1.6\times10^{-19}\cdot 0.23} $$
$$r=\color{red}{\bf 5.25\times 10^{-5}}\;\rm m$$
Now we need to find the pitch $p$ and we know that the horizontal velocity component of the electron is constant.
We also know that the pitch is the distance traveled during completing one circular path.
So,
$$p=v_xt=v\cos45^\circ \cdot \dfrac{2\pi r}{v\sin45^\circ }$$
$$p= \dfrac{2\pi r\cos45^\circ }{ \sin45^\circ }$$
Plugging the known;
$$p= \dfrac{2\pi \cdot 5.25\times 10^{-5}\cdot \cos45^\circ }{ \sin45^\circ }$$
$$p=\color{red}{\bf 3.3\times 10^{-4}}\;\rm m$$