Work on a proton in uniform magnetic field

[Elbobo]

Homework Statement

A magnetic field of 0.1 T forces a proton beam
of 1.5 mA to move in a circle of radius 0.1 m.
The plane of the circle is perpendicular to the
magnetic field.
Of the following, which is the best estimate
of the work W done by the magnetic field on
the protons during one complete orbit of the
circle?
1. W = 0 J
2. W = 10^22 J
3. W = 10^−22 J
4. W = 10^20 J
5. W = 10^−5 J

Homework Equations

F = qVB = ILB
F = mv^2 / r
C = 2 pi*r

The Attempt at a Solution

Since work is energy, I used the formula for kinetic energy...

W = 1/2 mv^2

F_centripetal = F_magnetic
mv^2 /r = qvB
v = rqB / m

W = 1/2 m (rqB / m)^2
W = 1/2 (1.6725 x 10^-27) * (0.1 * 1.602 x 10^-19 * 0.1 / 1.6725 x 10^-27) ^2
W = 7.67235874 x 10^-16 J

This isn't close to any of the answer choices. Can someone help me?

 

[LowlyPion]

The Force supplied by the B field is

F = qV X B (the cross product)

What is the direction of this force relative to the direction of the motion of the particle?

Isn't Work

W = F ⋅ D (the dot product)

 

[Elbobo]

I tried a multitude of ways, including that, and the order was between -13 and -17. The closest answer to that is choice 3, but that's wrong.

F = qvB
W = qvB * 2 pi r

(in part 2, I solved for v and found that it's approximately 10^6)

W = (1.602 x 10-19) (10^6) (0.1) * 2 pi * 0.1
W = 1.00656629 x 10^-14 J

Again, nothing close.

 

[Elbobo]

OR, does anyone think choice 5 is a typo, and it meant to be 10^-15?

My online system sometimes has problems/answers like that unfortunately...

 

[LowlyPion]

What angle is the B field and the Force making with the direction of motion?

 

[Elbobo]

Everything is at a right angle.

 

[LowlyPion]

So how much work does a force at right angle do again?

Work is F⋅D which is the dot product.

So what is the dot product of the Force with the direction of motion?

 

[Elbobo]

Ohhh, so it's zero. Gah, I was supposed to remember that detail from that chapter we learned months ago.

Thanks again for all your help, LowlyPion.