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Given magnetic field, particle charge, and force (vector)... Calculate velocity? |
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| Apr6-06, 08:27 PM | #1 |
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Given magnetic field, particle charge, and force (vector)... Calculate velocity?
A particle with charge -5 nC is moving in a uniform magnetic field B = -(1.2 T)k. The magnetic force on the particle is measured to be F = -(3.6*10^(-7) N)i + (7.6*10^-7)j. Calculate the x and y components of the particle's velocity.
F = q(V X B) (force equals charge multiplied by the cross product of V and B) Calculate the scalar product (dot product) v*F vx*Fx + vy*Fy + vz*Fz? What is the angle between v and F (in degrees)? F = qvB sin theta theta = arcsin [F/(qvB)]? |
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| Apr6-06, 10:14 PM | #2 |
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Fz is zero, right? So do you need to worry about Vz, or the angle between V and B? Or do you just need to worry about the component of V that is perpendicular to B? (Which would be in the xy plane only).
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| Apr7-06, 11:17 AM | #3 |
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I believe that I just need to worry about the component of V that is perpendicular to B (x and y components of V) since force exists in the x and y directions only.
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| Apr7-06, 12:59 PM | #4 |
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Given magnetic field, particle charge, and force (vector)... Calculate velocity?
I found equations for each of the x and y components:
V_x = (-F_y)/(q*B) V_x = -(-7.6E-7)/(-5E-6*-1.20) V_x = -0.127 m/s V_y = (-F_x)/(q*B) V_y = -(-3.6E-7)/(-5E-6*-1.2) V_y = 0.06 m/s But those answers aren't right... What am I doing wrong? Are the equations not right? Am I using incorrect values in correct equations? I'm so confused! |
| Apr7-06, 01:22 PM | #5 |
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Kay, nevermind...
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| Apr7-06, 01:31 PM | #6 |
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Just calculate the cross product q(vxB), assuming a certain variable for each component of the velocity, and equate the result to the force.
Btw, vxF is zero and it doesn't help much. |
| Apr7-06, 01:53 PM | #7 |
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Thanks. I found out my problem. I'm so used to converting micro 10^-6 that when I came across "nano," I used micro's conversion value.
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| Apr27-10, 08:56 PM | #8 |
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v*F being zero tells you that the angle between them is 90 degrees, so it does actually help
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