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!

 

[kyle8921]

a) Based on the given information, the speed of the ions can be calculated using the equation QV=(1/2)mv^2, where Q is the charge of the ion, V is the voltage, m is the mass of the ion, and v is the velocity. Plugging in the values, we get (1.6*10^-19C)(2000V)=(1/2)(6*1.67*10^-27kg)v^2. Solving for v, we get v=2.5*10^5m/s.

b) The electric field in the velocity selector plates can be calculated using the equation E=Bv, where B is the magnetic field and v is the velocity of the ions. Substituting the values, we get E=(0.020T)(2.5*10^5m/s)=5.0*10^4N/C.

c) The voltage in the velocity selector plates can be calculated using the equation V=Ed, where E is the electric field and d is the distance between the plates. Substituting the values, we get V=(5.0*10^4N/C)(0.05m)=2500V. The reason for the discrepancy in the calculations is most likely due to rounding errors or significant figures. It is important to carry out calculations with the appropriate number of significant figures to get accurate results.