Relationship between force and Velocity in Magnetic Fieldsby Typhon4ever Tags: fields, force, magnetic, relationship, velocity 

#1
Apr1713, 10:07 PM

P: 43

A negative particle is moving in a uniform magnetic field pointing in the negative k direction. The force on the particle is i and j. Find the x and y components of velocity. (I left out the numerical data in the question). I used F=q*v*B and in order to find the x component I used the F in the x direction with is wrong. You need to use the F in the y direction to find x component. Why?




#2
Apr1713, 10:30 PM

P: 2,493

The force on a charged particle acts perpendicular to the direction of travel and the B field following right hand rule for cross products of vectors:
F = q(E + v x B) where F, E and B are vector quantities and x means cross product. 



#3
Apr1713, 10:35 PM

P: 43





#4
Apr1713, 11:10 PM

P: 2,493

Relationship between force and Velocity in Magnetic Fieldsgives the magnitude of F. 



#5
Apr1713, 11:15 PM

P: 43

I'm confused. If there is a force in the i direction on the particle as well as a force in the j direction and we want the i and j velocities why don't we just use the corresponding forces in the corresponding directions? A force in the i direction will affect the i velocity won't it?




#6
Apr1713, 11:30 PM

P: 43

Hmm. I was using F=qvB sin(theta) but I don't know sin(theta). I should have used F=qv x B because I know then that F is perpendicular to v so I must choose the perpendicular force. Correct? Or are you still able to use the angle version.




#7
Apr1813, 12:32 AM

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PF Gold
P: 11,102

There are lots of ways to evaluate the cross product  u x v = uvsinθ is one of them. However, this relation only computes the magnitudes, the question is asking about directions. If you put the magnitudes equal to 1 for each vector you can find sinθ  but it is more convenient to evaluate the vector cross product directly. It is even easier to do it using the righthand rule. 



#8
Apr1813, 02:23 AM

P: 43





#9
Apr1813, 02:37 AM

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PF Gold
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I misspoke ... my apologies.
I got confused because you have not provided all the information given to you about the problem. You appear to have given us the force direction and magnitude, the magnetic field direction only, and only the sign of the charge. I assumed which more information was available to you without checking first. 



#10
Apr1813, 02:45 AM

P: 43

ok to be specific the charge is 5.00 nC, B=(1.28T)k, Magnetic F= (3.30×10^7 N)i+(7.60×10^−7 N)j. Does that change anything?




#11
Apr1813, 03:10 AM

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That changes the magnitude and direction of the force for starters... which changes the plane that the velocity is in. But it confirms what I thought  you have to do the vector math.
rewriting as vectors... $$\vec{F}=\begin{pmatrix}3.30\\7.60\\0\end{pmatrix}\times 10^{7}\text{N} \; ;\; \vec{B} = \begin{pmatrix}0\\0\\1.28\end{pmatrix}\text{T}\; ;\; \vec{v}=\begin{pmatrix}v_x\\v_y\\v_z\end{pmatrix}\text{m/s}\; ;\; q=5.00\times 10^{9}\text{C}\\ \vec{F}=q\vec{v}\times\vec{B}$$... do you know how to do a cross product? Note  you can only find the x and y components of the velocity. Fortunately, that is all they ask for. 


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