Verifying Coordinate System for Electric and Magnetic Forces

[mcastillo356]

Homework Statement

One proton experiences a force of $$1,10\cdot{10^{-13}}\;N$$ in the direction $$+y$$ due to a electric field
a) Calculate the direction and size of the electric field that causes the force
At a certain moment has a velocity of $$2\cdot{10^5}\;m/s$$
b) Calculate the magnetic field (size and direction) needed to apply so that the proton experienced a net force equal to zero
Data: $$q(p)=1,6\cdot{10^{-19}}\;C$$

Relevant Equations

$$\vec{E}=\dfrac{\vec{F_e}}{q}$$
Newton's second law
$$\vec{F_m}=q(\vec{v}\times{\vec{B}})$$

The attached file is the coordinate system I've used
a) $$\vec{E}=\dfrac{\vec{F_e}}{q}=\dfrac{1,10\cdot{10^{-13}}\hat{j}\;N}{1,6\cdot{10^{-19}}\;C}=6,88\cdot{10^5}\hat{j}\;N/C$$
b) $$\sum{\vec{F_{net}}}=\vec{0}=\vec{F_e}+\vec{F_m}$$
$$-6,88\cdot{10^5}\hat{j}\;N=q(\vec{v}\times{\vec{B}})=q\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\v_x&v_y&v_z\\B_x&B_y&B_z\end{vmatrix}=q\left[(v_yB_z-v_zB_y)\hat{i}-(v_xB_z-v_zB_x)\hat{j}+(v_xB_y-v_yB_x)\hat{k}\right]=-qv_xB_z\hat{j}$$
The direction of $$\vec{B}$$ is $$+z$$
So $$\dfrac{6,88\cdot{10^5}\hat{j}\;N}{q\cdot{v_x}}\hat{k}=\vec{B}$$
$$2,15\times{10^{19}}\hat{k}T=\vec{B}$$
Have I done wright?

 

[berkeman]

At a certain moment has a velocity of $$2\cdot{10^5}\;m/s$$

It looks like you have only listed the scalar speed, not the vector velocity with direction. Was the direction of the proton's motion given in the problem?

 

[hutchphd]

Check the units of E in your penultimate equation. You are off by ~$10^{19}$
Notice this balance holds only for a proton at a given speed (velocity actually @berkeman ) and the arrangement is very useful as a velocity selector for a charged particle beam (used in mass spectrometers often)

 

[TSny]

 

[mcastillo356]

Hi everybody. berkeman, hutchphd, TSny:
berkeman, the velocity they provide is only the given in the statement

And this other?:
$$-F_e\hat{j}\;N=-q\cdot{v_\color{red}x}\cdot{B_z}\hat{j}\;N$$
$$-F_e=-q\cdot{v_\color{red}x}\cdot{B_z}$$
$$-F_y=-q\cdot{v_\color{red}x}\cdot{B_z}$$
$$B_z=\dfrac{F_y}{q\cdot{v_\color{red}x}}=3,44\;T$$
$$\vec{B}=3,44\;T\hat{k}$$

Salutes to everybody