How Much Is a Charged Bullet Deflected by Earth's Magnetic Field?

[ts174ab2]

A 3.10 bullet moves with a speed of 165 m/s perpendicular to the Earth's magnetic field of 5.00x10^-5.

If the bullet possesses a net charge of 1.36×10−8C , by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.35km ?

Could anyone give me a hand (ie.equation)?
I have no idea how to start the question...

 

[dynamicsolo]

You can use the information available to calculate the magnetic force on the bullet, from F = qv x B. This will give you the magnitude of the force, from which you can find an acceleration. Keep in mind that, since this is a magnetic field, the force and acceleration you have found are centripetal, so the bullet would be moving on a circular arc.

This means that the 1.35 km. is measured along the arc. You can also work out the radius of the circle, knowing the relationship for centripetal force. This will tell you the angle on the circle through which the bullet moves. Geometry will then give you the amount by which the bullet will have been deflected from the straight line it was traveling on initially.

 

[ts174ab2]

You can use the information available to calculate the magnetic force on the bullet, from F = qv x B. This will give you the magnitude of the force, from which you can find an acceleration. Keep in mind that, since this is a magnetic field, the force and acceleration you have found are centripetal, so the bullet would be moving on a circular arc.

This means that the 1.35 km. is measured along the arc. You can also work out the radius of the circle, knowing the relationship for centripetal force. This will tell you the angle on the circle through which the bullet moves. Geometry will then give you the amount by which the bullet will have been deflected from the straight line it was traveling on initially.

sorry, I didn't understand the second part.
Could you give some detail steps? thanks!

 

[dynamicsolo]

It is really preferable to conduct this work in the Forum, rather than by private-messaging. For one thing, not having replies posted here makes it look like the thread is inactive; other helpers will assume your question is not being answered and will start posting when it may not actually be necessary for them to do so.

To continue from your PMs, you got a radius for the circular path of the bullet which is rather gigantic:

r = mv/qB = 7.52·10^11 m. (far larger than the Earth itself, but smaller than its orbit!)

Maybe this isn't surprising, since the charge is rather small and the bullet is enormously massive by charged particle standards.

You also found that the angle that the bullet travels along the circle is

theta = arclength/radius = 1350 m./7.52·10^11 m.
= 1.80·10^-9 radians .

Since the circle is so vast compared to the 1.35 km the bullet travels on it, we can pretty well take the path of the bullet to be a straight line, but now turned 1.8·10^-9 radians from its original direction.

You can then use the "skinny-triangle" approximation to find how much the bullet has been deflected from where it would have arrived after moving the 1350 meters. (It's mighty dang little: this is probably the point of the problem...)