What is the Direction of Magnetic Force on a Moving Electron?

[Jimbob999]

Homework Statement

An electron (charge = –1.6 x 10^–19 C) is moving at 3.0 x 10^5 m/s in the positive x direction. A magnetic field of 0.80 T is in the positive z direction. The magnetic force on the electron is:

http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

a) 0 N

b) 3.8 x 10^–14 N in the positive z direction

c) 3.8 x 10^–14 N in the negative z direction

d) 3.8 x 10^–14 N in the positive y direction

e) 3.8 x 10^–14 N in the negative y direction

Homework Equations

F(b) =q(vxb)

The Attempt at a Solution

Force is 3.8x10^-14

In terms of direction, I was told that it was the positive y direction, but isn't the charge negative so it should be the negative y direction?

 

[berkeman]

I get it in the positive y direction, because X x Y = Z. Can you post a diagram?

 

[berkeman]

BTW, why are all of the answers wrong?

 

[Jimbob999]

BTW, why are all of the answers wrong?

They are all wrong? I got the same force as the other force answers...

I know X x Y = Z, but from my textbook "If q is negative, then the force and cross product have opposite signs and thus must be in opposite directions. "
Shouldn't this then reverse it from positive y direction to negative y direction?

 

[TSny]

I know X x Y = Z, ...

But what is X x Z ?

 

[Jimbob999]

But what is X x Z ?

You mean qv x F? I am not sure I follow?

 

[Jimbob999]

BTW, why are all of the answers wrong?

F = –1.6 x 10^–19 (3.0 x 10^5 x0.8)
F =-3.84 x 10^–14

Admittedly I get a negative answer, but at least it matches numerically.

 

[Qwertywerty]

Admittedly I get a negative answer, but at least it matches numerically.

I think you mean it's magnitude matches .

The direction - assuming z - axis leaves the plane , how are you getting force in the -ve y direction ?
Velocity vector's cross product with the magnetic field is in the -ve y direction . So , a -ve charge implies the opposite of this , i.e. , in the +ve y direction .

Hope this helps .

 

[Jimbob999]

I think you mean it's magnitude matches .

The direction - assuming z - axis leaves the plane , how are you getting force in the -ve y direction ?
Velocity vector's cross product with the magnetic field is in the -ve y direction . So , a -ve charge implies the opposite of this , i.e. , in the +ve y direction .

Hope this helps .

Again I would get the +ve y direction answer as you do, except the part in the textbook that confuses me is this: 'If q is negative, then the force and cross product have opposite signs and thus must be in opposite directions' q is negative isn't it? Thus the positive y becomes negative y. Oh boy how lost am I now...

 

[Qwertywerty]

Again I would get the +ve y direction answer as you do, except the part in the textbook that confuses me is this: 'If q is negative, then the force and cross product have opposite signs and thus must be in opposite directions' q is negative isn't it?

That's what I have said . v×B ( in vector form ) is towards -ve y direction , but electron charge is -ve , so force is in the +ve direction .

 

[Jimbob999]

That's what I have said . v×B ( in vector form ) is towards -ve y direction , but electron charge is -ve , so force is in the +ve direction .

Ah ok, I think I get that now.

So am I correct in concluding firstly that F = 3.84 x 10^–14 or as berkeman hinted at, is that incorrect?

 

[Qwertywerty]

Although you'll have to check the magnitude .