- #1

Kruum

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## Homework Statement

http://www.aijaa.com/img/b/00125/3627010.jpg

http://www.aijaa.com/img/b/00758/3627011.jpg

## Homework Equations

[tex]E_k=\frac{1}{2}mv^2[/tex]

[tex]\vec{F}=m\frac{\vec{v}^2}{r}[/tex]

[itex]\vec{F}=q\vec{v} \times \vec{B}[/itex]

[tex]\frac{v}{r}=\frac{2\pi}{T}=\omega[/tex]

## The Attempt at a Solution

The only difficulty in problem 2 is the unit of the mass of the proton. I can't use any info that's not given in the problem or I don't know by heart. I know the mass of the proton is about 1u, but I don't know how to convert it to kg. First I solve [tex]E_k[/tex] for [tex]v=\sqrt{\frac{2E_k}{m}[/tex] and then by setting the forces equal I get [tex]r=\frac{mv}{qB}[/tex]. Now here's where I need kg instead of u - or am I missing something.

The thing that gets me in number 5, is the way the magnetic field is given. If we set the Earth's surface along the xy-plane so that the positive y-axis points towards north and the positive x-axis towards east. Then [tex]\vec{B_h}=18\mu T\vec{j}[/tex]. But is [tex]\vec{B_v}=-54\mu T\vec{j}[/tex]? I think in that case they would have given the magnetic field as "[tex]B=36\mu T[/tex] to south". So they only option left is [tex]\vec{B_v}=-54\mu T\vec{k}[/tex]. Which one is it and why? From here the problem is pretty straight forward: I just need to find the cross product of the vectors and calculate its length.

I'm sorry for the non-native English, but ask if you can't understand something I've written.

Edit: Some typos...

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