Electricity and Magnetism in a veloctiy selector

[Mr. Snookums]

Lithium-6 ions (mass is 6x a proton mass) are charged Li+ ions (one extra proton). They are accelerated by 2000 volts and passed through a velocity selector with a magnetic field of 0.020T into the page. The ions follow a straight path through the selector and are then deflected onto a piece of film. The velocity selector plates are separated by 5.0cm.

Find:

a)the speed of the ions.

This is simple enough. QV=(1/2)mv^2, plug in the mass, charge and voltage and solve for velocity. This gives me 2.5*10^5m/s.

b)Electric field in the velocity selector plates.

QvB=QE, so E=Bv=(2.5*10^5)(0.020)=5.0*10^3N

However, I am off by a factor of ten. The answer is 5.0*10^4. Why is this?

c)Voltage in the velocity selector plates.

I know that the magnetic and electric forces are balanced. The electric force pushes down, the magnetic force pushes up.

For electric force, F=QE, so am I right in saying that F=QV/d?

Then I would get vB=V/d. This was my reasoning, but when I plug in the numbers, I do not get the correct answer. How would I go about solving this one?

 

[al_201314]

Interesting I get the same answer as you do. Is the answer definitely correct? If yes we got to wait for someone more experienced!