HELP: Charges in Magnetic Field Question! (1 Viewer)

[iheartOJ]

Hi,

Can anyone show me how to do this?

A single positively charged particle, mass 4.6x10^-27 kg enters a 4.0T magnetic field into the page at 2.5 x 10^6 ms^-1. The field covers an area of 0.1m x 0.1m. It enters at 90o, 0.05m up from the bottom left corner. Calculate the force on the particle when it is in the field.

ANSWER:1.8 x10^-12 N

Thanks!

 

[YoloStudent]

F=qvbsin(angle)?

q= 1.6x10^-19
v = 2.5x10^6
b=4
angle = 90

answer is close... not sure if information wrong or something wrong with me

 

[seventhroot]

when it says "single positively charged" are we assuming a proton?

 

[seventhroot]

I am thinking we have to equate an expression with (mv^2)/r

 

[YoloStudent]

Perhaps we're using F=qvbSin(angle) with F=mv^2/r and playing around with the equations to make force the subject?

 

[seventhroot]

Perhaps we're using F=qvbSin(angle) with F=mv^2/r and playing around with the equations to make force the subject?

every time I do something like that; the force cancels lol

 

[MaccaFacta]

This is the correct method. But iheartOJ might have left something out when they retyped the question - remember force is a vector. So you might need to state both the size and the direction in order to fully answer the question. A positively charged object travelling into the page will experience a force which is at right angles to the plane which contains its velocity vector and the magnetic field vector.