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## Homework Statement

So there's a rectangular region in the xy plane subjected to a magnetic field B with strength 0.06T pointing in the -z. A proton moving in the xy plane enters the region with velocity v such that the velocity components are initially vx = v*cos(24), vy = v*sin(24). It leaves the rectangular region at a point 0.4m above (as in yf = yi + 0.4m, xf = xi) its entry with the same vy and opposite vx. What is v?

## Homework Equations

F = qv x B, qvB = mv^2/r

## The Attempt at a Solution

I figured I could just use F = qb x B to find the force in the y direction and then divided by the mass and integrated twice and set it equal to 0.4m, then solve for v, like

F/m = a = qvBcos(∅)/m

so=

x = 0.4 = ∫∫vqBcos(∅)/m d∅^2 + ∫vsin(θ)d∅ for ∅ = 24 to 154

then solve for v. The first integration is due to the B field and the second is just integrating the initial vy.

But I'm off by more than a couple orders of magnitude....

Any help would be appreciated! I suspect that my way isn't the easiest way, but if anyone could explain the flaw in my approach anyway that would be awesome.

Thanks for the help!

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