# Uniform Magnetic Field

Discussion in 'MCAT Study Question Q&A' started by mybubbles627, Apr 12, 2012.

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1. ### mybubbles627ASA Member 5+ Year Member

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May 10, 2011
West Coast
A proton and an electron are traveling in the uniform magnetic field with identical velocities. If the movement of both particles is perpendicular to the magnetic field lines, which of the following is NOT true?

I. The acceleration of the proton is greater than the acceleration of the electron

II. The proton will experience a greater kinetic energy change than the electron

III. The magnetic force on both particles is 0

A. III only
B. I and III only
C. II and III only
D. I, II, and III

The answer is D. I thought the answer was B because KE = 1/2mv^2 so I figured that the proton would have a greater kinetic energy.

The explanation is that the force doesn't do work and therefore can't change the kinetic energy of either particle.

Can someone explain this in another way to me please? I get why I and III are false, but I don't get why II is false.

3. ### syoungMS-3 5+ Year Member

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Jan 2, 2011
Over the rainbow
MDApps:
Well to have a change in KE, the KE must turn to another form of energy (either potential or something else). We can assume these are conservative forces, so no friction is done.
Well F=qvB, so the force on BOTH are the same (they have same magnitude of charge but different sign) and same velocity and same B (mag field), so they will experience the same F (which is nonzero).
As they feel the same force, they will get deflected the same amount. Whatever kinetic energy change is felt by the + will be felt by the -. Now the + may not get deflected as much because of its larger size compared to the -. But they will feel the same "change" in kinetic energy.

4. ### Temperature101BannedBanned 2+ Year Member

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May 26, 2011
D is correect since the direction of the force is perpendicular to the velocity of both proton and electron, there is no work done on them (KE= 0).. Remember that: Any question on magnetic filed that ask for KE or Work is ZERO.

5. ### syoungMS-3 5+ Year Member

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Jan 2, 2011
Over the rainbow
MDApps:

OOps forgot about the cross product of F=qvB! darn... F is perpindicular to velocity! darn darn darn

6. ### mybubbles627ASA Member 5+ Year Member

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May 10, 2011
West Coast
Are you using W = Fdcos(theta) and since they are perpendicular cos(theta) is 0 therefore the W = 0 therefore the KE= 0?

7. ### mybubbles627ASA Member 5+ Year Member

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May 10, 2011
West Coast
So basically the KE of the proton IS greater than the KE of the electron BUT because it asks for the CHANGE in KE the change is the same? Not really following this.....

3,489
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May 26, 2011
YES and YES

9. ### milski1K member 5+ Year Member

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Dec 30, 2009
Where the rain grows
There is no work done on either of them, thus the change in KE for each of them is 0. Yes, the proton has higher KE but since neither changes KE, the change is the same for both.

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10. ### mybubbles627ASA Member 5+ Year Member

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May 10, 2011
West Coast
But HOW do you KNOW no work is done? I don't fully understand that because I know you can't always use the W = Fdcos(theta) equation so how do you KNOW you can in this situation to get that no work is done?

Should I just memorize no work is EVER done for uniform magnetic field problems? Is that because the proton/electron moves in a circle so the start point and end point are the same?

11. ### milski1K member 5+ Year Member

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Dec 30, 2009
Where the rain grows
Careful there. The work is zero only if the magnetic force is perpendicular to the velocity. It's easy to come up with problems where that is not true.

Also, the KE is not zero (unless the particles are at rest), only the change in KE.

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12. ### milski1K member 5+ Year Member

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Dec 30, 2009
Where the rain grows
Of course you can use the formula - it just has to be for an infinitely small displacement. Since you know that the force will always stay perpendicular to the velocity, it will also be perpendicular to the displacement. Then the cosine will be zero and no work be done.

You could memorize it but be careful how you apply it - it is correct only for uniform field perpendicular to the velocity and only when there are no other forces acting.

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3,489
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May 26, 2011
Got you....