Direction of magnetic field and moving particle

  • #1
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Homework Statement


A particle of mass 0.195 g carries a charge of -2.50 x 10-8C. The particle is given an initial horizontal velocity that is due north and has magnitude 4.00 x 104 m/s. What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

Homework Equations


F = |q|vperpendicularB = |q|vBsin##\theta##
##\vec F = q\vec v x \vec B##

The Attempt at a Solution


For magnitude,

F = mg = |q|vB

##(.195 x 10^{-3})(9.81) = (2.50*10^{-8})(4.00*10^4)(B)##
##B = \frac {.195 x 10^{-3})(9.81)} (2.50*10^{-8})(4.00*10^4)(B)##
##B = 1.91 T##

the magnetic field is always perpendicular to the velocity, and since velocity is north, the magnetic field is east or west. But how do I choose east or west?
 

Answers and Replies

  • #2
TSny
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F = mg = |q|vB

##(.195 x 10^{-3})(9.81) = (2.50*10^{-8})(4.00*10^4)(B)##
##B = \frac {.195 x 10^{-3})(9.81)} (2.50*10^{-8})(4.00*10^4)(B)##
##B = 1.91 T##
Looks good.

the magnetic field is always perpendicular to the velocity
In general, the B field is not required to be perpendicular to the velocity. In this problem, there is a particular reason why you want B to be perpendicular to the velocity. Can you state the reason?

since velocity is north, the magnetic field is east or west.
East and west are not the only directions that are perpendicular to north.

But how do I choose east or west?
The magnetic force must be in what direction? Choose the direction of B so that the force will have this direction.
 
  • #3
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oops..I read the book wrong.
It says the direction of the force is always perpendicular to the velocity vector and B field. And I realize that if the z-axis is north, then any vector on the x-y plane is perpendicular to north.

By making the angle 90 degrees, we get the maximum magnitude of the force. So should I have chosen an angle like 1 degrees to find the minimum magnitude, since the question asks for the minimum?

I need the magnetic force to cancel out the force of gravity.
so ## \vec F_g + \vec F_b = 0##
##\vec F_g = (9.81)(.195*10^{-3}) > 0##
so ##\vec F_b < 0##

##\vec F_b = q\vec v x \vec B##
##\vec F_b = (-2.50 * 10^{-8}(4.00 *10^4) x (\pm 1.91) < 0##
So B must be positive. So the B field is in the positive direction..
 
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  • #4
TSny
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oops..I read the book wrong.
It says the direction of the force is always perpendicular to the velocity vector and B field. And I realize that if the z-axis is north, then any vector on the x-y plane is perpendicular to north.
Yes, good.

By making the angle 90 degrees, we get the maximum magnitude of the force. So should I have chosen an angle like 1 degrees to find the minimum magnitude, since the question asks for the minimum?
You know the magnitude of the magnetic force that is required. You want to get that much force using the smallest possible magnitude of B field. So, you should think about the angle that will allow you to get the required amount of force using the smallest B field.

I need the magnetic force to cancel out the force of gravity.
so ## F_g + F_b = 0##
##F_g = (9.81)(.195*10^{-3}) > 0##
so ##F_b < 0##

##F_b = \frac {(.195 * 10^{-3})(9.81)} {4.00 * 10^4)(B)}##
##F_b = \frac {+} {(+)(B)}##
So B must be negative. So the B field is west?
You can't get the direction of B by just using + and - signs to indicate direction. Use the right hand rule for relating the direction of force and the direction of the B field.
 
  • #5
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Ok I think i get it,

We if ##sin(\theta)## gets smaller, B has to get bigger because q is constant so B or v would have to get bigger. So by making ##sin(\theta)## as big as possible, B and v will be as small as possible.

My attempt at right hand rule: thumb points up toward ceiling(velocity); I turned my palm toward the ceiling(magnetic force); fingers point away from me, so they are pointing westward, so the direction of the B field is west.

Just to confirm.. my thumb is now pointing right. So this means "right" represents North and my fingers represent West.
 
  • #6
TSny
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I think you almost have it, except don't forget that your point charge is negative.
 
  • #7
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I think you almost have it, except don't forget that your point charge is negative.

The magnitude of ##F_b## is positive.. but since there was a negative point charge, the direction of ##F_b## is down.. so doing right hand rule, my fingers point East.. so the B field is in the East direction.
 
  • #8
TSny
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The magnitude of ##F_b## is positive.. but since there was a negative point charge, the direction of ##F_b## is down.. so doing right hand rule, my fingers point East
I'm not sure I follow your statements here.

.. so the B field is in the East direction.
Yes.
 
  • #9
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I'm not sure I follow your statements here.

What I meant is, I had the force direction as up. toward the ceiling. So when I did the right hand rule, my palm was facing toward the ceiling.
I did this with the charge as positive. But the charge is negative, so I read the force is pointing in the opposite direction. So then I put my palm facing toward the floor, and that made my fingers point East.
 
  • #10
TSny
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What I meant is, I had the force direction as up. toward the ceiling. So when I did the right hand rule, my palm was facing toward the ceiling.
I did this with the charge as positive. But the charge is negative, so I read the force is pointing in the opposite direction. So then I put my palm facing toward the floor, and that made my fingers point East.
OK. Sounds good.
 
  • #11
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OK. Sounds good.

Thanks for your help
 

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