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Electromagnetism - right hand rule

  1. Jul 30, 2007 #1
    1. The problem statement, all variables and given/known data
    A proton travelling at a speed of 3.0 x 10^6 m/s travels through a magnetic field of strength 3.0 x 10^-3 T, making an angle of 45 degrees with the magnetic lines of force. What force acts on the proton?

    2. Relevant equations

    F = QvB

    3. The attempt at a solution

    I arrived at F = 1.44 x 10^-15N by simply plugging in. Then using the 45 degrees, I assume the force I found is the hypotenuse so I solve for an adjecnt using trig to get the correct answer of 1.0 x 10^-15 N but I dont understand this conceptually. Why is the force at 45 degrees weaker, and why am I solving it this way?

    Thanks a lot
  2. jcsd
  3. Jul 30, 2007 #2
    The correct Lorentz force equation is

    [tex] \mathbf{F} = Q \mathbf{v} \times \mathbf{B} [/tex]

    where [itex]\mathbf{v} \times \mathbf{B} [/itex] stands for the cross-product of two vectors. Since you know the angle [itex] \alpha = 45 [/itex] degrees between [itex]\mathbf{v} [/itex] and [itex]\mathbf{B} [/itex], you can use standard formula for the length of the cross-product vector

    [tex] | \mathbf{v} \times \mathbf{B}| = |\mathbf{v}| |\mathbf{B}| \sin \alpha [/tex]

    to obtain the magnitude of the force [itex]|\mathbf{F}| [/itex].

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