# Homework Help: Magnetic Force on a Particle

1. Aug 14, 2004

### RunsWithKnives

ok I'm stuck here...

At a particular point in time, an electron is moving with a velocity of
v=(2.0*10^6 m/s)i + (3.0*10^6 m/s)j

in a constant magnetic field with strength
e=(0.030T)i - (0.15T)j

What is the magnitude of the magnetic force on the electron at that moment? this i found by taking the cross product and multiplying by charge of an electron 6.246e-14N

The electron moves into an area where an electric field is used to bring its velocity to zero in 4.5 X 10-2 seconds. At the moment when the velocity of the particle is zero, what is the magnitude of the magnetic force on the particle?

this is where I'm stuck... i thought that vf = vi + at would give me the acceleration which I could then multiply by the mass of an electron (F=ma)... which would give me F... I know I'm probably missing something elementary here

so I got the magnitude of acceleration by taking the squares of both components of velocity and than the square root... as in sqrt(i^2 + j^2) = v
which gave me 1.3e13 m/s

so then I got 0 m/s = 1.3e13 m/s + a (4.5e-2 s)

a=-2.888e14 m/s^2

F= (9.110e-31 kg)*(2.888e14 m/s^2)

F=2.632e-16 N

so what am i missing?? thanks :grumpy:

2. Aug 14, 2004

### Gokul43201

Staff Emeritus
Haven't read it all but I think you forgot to take the square root here...so really,
|v| = 3.6 e6 m/s

3. Aug 15, 2004

### RunsWithKnives

stupid mistake :yuck: ... changes my answer to 7.299e-23 N... which still isn't right.... any other ideas ??

edit: I'm guessing that I'm missing something obvious... I'm supposed to do this using vectors since F = q VxB ..... right?? this was how I did the first part... I'll try again in the morning.. too tired to think :zzz:

Last edited: Aug 15, 2004
4. Aug 15, 2004

### Gokul43201

Staff Emeritus
Either you've not copied the question down correctly, OR there are more sub-questions following this OR it's a trick question.

Look at it again : "What is the magnetic force when the velocity is zero ?"

5. Aug 15, 2004

### maverick280857

Hi

Does the region where the electric field declerates the particle, also include the magnetic field? Otherwise, the magnetic force + direction at the point of entry into this new region can be computed using the Lorentz force definition. The electric field must have a magnitude and direction so as to bring down the velocity to zero.

Now at the point the velocity becomes zero, the acceleration is nonzero. But now your answer depends on whether B is operating in that region or not. If it is not operating, then qv(cross)B is the magnetic force and it is zero irrespective of whether v is zero or not. Note that the total force is always qv(cross)B + qE where E is the electric field (the trivial case is that one of the fields is zero, when the Lorentz force reduces to either the magnetic force expression or the electric force one).

Cheers
Vivek

6. Aug 15, 2004

### RunsWithKnives

it was a trick question... I am soo angry right now... there is no magnetic force unless the particle has a velocity..... AHHHHHHHHHHHHHHHHHHH
the answer is zero!!!!!!!!

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