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Determining the Magnetic Force on a Moving Charged Particle

  • Thread starter robera1
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  • #1
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Homework Statement


If the magnetic field of the wire is 2.5×10^−4 and the electron moves at 1.0×10^7 , what is the magnitude of the force exerted on the electron?

Homework Equations


F=qvBsin(theta)

The Attempt at a Solution


Sin(theta) = sin90 = 1
q = -1
v = 1e7
B = 2.5e-4
So, (-1)x(1e7)x(2.5e-4) = -2500, but that is not the right answer. What am I doing wrong?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
diazona
Homework Helper
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2.5×10^−4 what? Tesla? Gauss? What about 1.0×10^7? Where are the electron and wire in space? How are they moving? It's impossible to figure out what you might have been doing wrong without knowing the full problem.
 
  • #3
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Oh, okay. This is all of the information they give...

Learning Goal: To practice Tactics Box 24.2 Determining the magnetic force on a moving charged particle.

When a particle of charge moves with a velocity in a magnetic field , the particle is acted upon by a force exerted by the magnetic field. To find the direction and magnitude of this force, follow the steps in the following Tactics Box. Keep in mind that the right-hand rule for forces shown in step 2 gives the direction of the force on a positive charge. For a negative charge, the force will be in the opposite direction.

TACTICS BOX 24.2 Determining the magnetic force on a moving charged particle
Note the direction of v and B, and find the angle [alpha] between them.
The force is perpendicular to the plane containing v and B. The direction of F is given by the right-hand rule.
The magnitude of the force is given by F = qvBsin[alpha]

Part C
If the magnetic field of the wire is 2.5×10^−4 T and the electron moves at 1.0×10^7 m/s, what is the magnitude F of the force exerted on the electron?
 
  • #4
diazona
Homework Helper
2,175
6
Okay... well, I'm guessing there's a diagram or something that goes along with that, that shows the wire, the electron, and the angle between the electron's velocity and the magnetic field? Not that it matters, as long as you've got the right angle (90 degrees).

Anyway, to continue one point from the previous post: you must keep track of units in your calculations. Think about this: what are the units for q, v, B? what unit is your previous answer (-2500) in?
 
  • #5
22
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Well, the answer is supposed to be in N.
And, since I am multiplying C, T, and m/s, then the -2500 should be in N also
 
  • #6
rock.freak667
Homework Helper
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31

Homework Statement


If the magnetic field of the wire is 2.5×10^−4 and the electron moves at 1.0×10^7 , what is the magnitude of the force exerted on the electron?

Homework Equations


F=qvBsin(theta)

The Attempt at a Solution


Sin(theta) = sin90 = 1
q = -1
v = 1e7
B = 2.5e-4
So, (-1)x(1e7)x(2.5e-4) = -2500, but that is not the right answer. What am I doing wrong?
The charge on an electron is [itex]-1.60\times 10^{-19}C[/itex]
 
  • #7
8
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Consider your charge. You multiplied by (-1), but what are the units of that? Units of fundamental charge, which is NOT 1 C.

EDIT: Lol, What he said ^
 
  • #8
22
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Fantastic! I got the answer... thanks!!
 
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