Deflection and Magnetic Fields in Thomson's Experiment

[howin]

Homework Statement

Calculate the magnetic field required to produce a deflection of .20 radians in Thomson's experiment. Assume the potential across the deflection plate is u=200V, the length of the deflection plates is x=5.0 cm, and their separation d=1.5cm. Compare this value of B to the Earth's magnetic field (B=.5 G = .5 * 10^-4 T).

Homework Equations

e/m(e) = u theta / B^2 * x * d

e/m(e) = 1.7588 * 10^11 C/kg

The Attempt at a Solution

1.7588 * 10^11 C/kg = (200V) (.2 rad) / (B^2)(.05m)(.015m)

B^2 = (1.516*10^-6 T^2)(.2 rad)

Now I can't figure out how to convert theta from radians to make this make sense. Do I just multiply by .2? Then I get that B=.55*10^-3 T, but I think the right answer is closer to the Earth's magnetic field.

 

[Jhero]

Dear fellow scientist,

To calculate the magnetic field required for a deflection of 0.20 radians in Thomson's experiment, we first need to rearrange the equation e/m(e) = u theta / B^2 * x * d to solve for B. This gives us B = √(u theta / e/m(e) * x * d).

Next, we can plug in the given values: u = 200V, theta = 0.20 radians, e/m(e) = 1.7588 * 10^11 C/kg, x = 0.05m, and d = 0.015m.

This gives us B = √((200V)(0.20 rad) / (1.7588 * 10^11 C/kg)(0.05m)(0.015m)) = 0.00203 T.

Comparing this value to the Earth's magnetic field of 0.5 * 10^-4 T, we can see that the required magnetic field for the deflection in Thomson's experiment is significantly stronger.

I hope this helps with your calculations. Keep up the good work!