Deflection of bullet traveling perpendicular to Earth's magnetic field

Question

A 0.0034kg bullet moves with a speed of 160 m/s perpendicular to the Earth's magnetic field of 5E-5 T. If the bullet has a net charge of 13.5E-9 C, by what distance is it deflected due to the magnetic field after it has traveled 1000m?

First, I found the radius of the circle via: r = (mv/qB) = 8.1E11

Next, I found theta using the arc length of the circle traveled by the bullet divided by the radius of that circle: theta = arc/radius = 1000/8.1E11

Then, to find the distance it's deflected, I used the triangle approach: tan(theta) = x/1000

However, my answer is incorrect. Any ideas would be great,

thanks in advance.

Answer 1

Maybe I am being stupid here, but I tend to see things in a different way to most people so I may not be answering this question from the right frame of mind, do forgive me if that is the case:

My response is, why would it be deflected at all? Bear with me here.

If the bullet is traveling PERPENDICULAR to the Earth's magnetic field, isn't that somewhat like a kite floating directly against the wind? The kite will be slowed down, but will not be blown in any differing directions due to the fact that the wind is hitting it dead on, and it is still being pulled forwards in the same direction.

However, metals are not repelled by magnets, only attracted. So if that is the case, then according to my limited understanding, if a bullet was flying perpendicular to a magnetic field it will not be repelled, and therefore will not be deflected at all.

If I am being stupid and I got the field idea backwards, then atthe very least the bullet will be pulled forwards a little faster, but that surely is not a deflection, since a deflection would mean it was thrown off course in either direction, irrespective of its inertia, right?

Or maybe I am just splitting hairs on that one?

The Founder

Answer 2

You're almost there - the vertical deflection (we'll just assume the dimension of deflection is verticle) would be r - r cos theta = r ( 1 - cos theta ). As small as theta is, I'd be tempted to just use the first few terms of the series expansion for cosine. And I'd draw a picture if I could :)