A velocity selector has its electric field directed vertically downward, with ma

Question

A velocity selector has its electric field directed vertically downward, with magnitude 130 N/C .

Find the magnitude of the horizontal magnetic field required to select charged particles moving at 340 m/s .

This velocity selector is being used on protons. Compute explicitly the magnitude of the electric force.

Compute explicitly the magnitude of the magnetic force.

(I have entered the answer with Newtons as the units and it is saying "Enter your answer using dimensions of magnetic flux density." Can anyone help me with this?

I will leave good rating. Thanks

Explanation / Answer

Velocity selectors will select particles moving with a velocity equal to the electric field divided by the magnetic field.
v = E/B ========> B = E/v = 130/340 = 0.382 T
The electric force acting on the proton is Fe = e*E = 1.6*10^-19*130 = 208*10^-19 N
The magnetic force acting on the proton is Fm = e*v*B = 2.08*10^-17 N

Where e is the proton charge.

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