Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
From the given graph, we can see that:
- $E=( -10^5\;\hat j)\;\rm V/m$
- $v= ( 10^6\;\hat i)\;\rm m/s$
- $\sum F=0\;\rm N$
Since the electron will move without deflecting, the net force exerted on it is zero, as mentioned above.
Recalling that the net force exerted on a moving charge in an electromagnetic field is given by
$$\sum F=q(\vec E+\vec v\times \vec B)$$
$$0=q(\vec E+\vec v\times \vec B)$$
Thus,
$$\vec E=-\vec v\times \vec B$$
$$\vec B=\dfrac{\vec E}{-\vec v}$$
Thus,
We can see that the electric force is downward, so the magnetic force is upward. So, according to the right-hand rule, the direction of the magnetic field is into the page.
Plug the known;
$$\vec B=\dfrac{10^5}{10^6 }=-(\color{red}{\bf 0.1}\;\hat k)\;\rm T$$
$$\color{blue}{\bf [b]}$$
We need to use Galilean fields transformation equations,
$$\left.\begin{matrix}
&\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A \\
& \vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}\\
\end{matrix}\right\}\tag 1$$
Assuming that B is the proton's frame.
So,
$$\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A$$
Plug the known;
$$\vec E_B=( -10^5\;\hat j)+ ( 10^6\;\hat i)\times (-0.1\;\hat k)$$
$$\vec E_B=( -10^5\;\hat j)+ ( 10^5\;\hat j) $$
$$\vec E_B=\color{red}{\bf 0}\;\rm V/m$$
And,
$$\vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}$$
Plug the known;
$$\vec B_B=(-0.1\;\hat k)-\dfrac{( 10^6\;\hat i)\times ( -10^5\;\hat j)}{(3\times 10^8)^2}=-0.099\;\hat k$$
$$\vec B_B=(-\color{red}{\bf 0.1}\;\hat k) \;\rm T$$
$$\color{blue}{\bf [c]}$$
The experimenter on the proton's frame sees no electric force since $E_B=0$, and sees no magnetic force since the proton is at rest in his frame.