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A proton travels with a speed of 3.39×106 m/s
at an angle of 56.8
◦ with a magnetic field of
0.305 T pointed in the y direction.
The charge of proton is 1.60218 × 10−19 C.
What is the magnitude of the magnetic
force on the proton?
Answer in units of N

  • physics -

    F = Q V x B
    where the x denotes a vector cross product.

    The magnitude of F is Q*V*B sin56.8
    = (1.602*10^-19)*(3.39*10^6)*(0.305)*(0.837)
    = _______ Newtons

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