?[T] The force vector F acting on a proton with an electric charge of \(1.6 \times

Chapter 2, Problem 240

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[T] The force vector F acting on a proton with an electric charge of \(1.6 \times 10^{-19} \mathrm{C}\) moving in a magnetic field B where the velocity vector v is given by \(\mathbf{F}=1.6 \times 10^{-19}(\mathbf{v} \times \mathbf{B})\) (here, v is expressed in meters per second, B in T, and F in N). If the magnitude of force F acting on a proton is \(5.9 \times 10^{-17}\) N and the proton is moving at the speed of 300 m/sec in magnetic field B of magnitude 2.4 T, find the angle between velocity vector v of the proton and magnetic field B. Express the answer in degrees rounded to the nearest integer..

Text Transcription:

1.6 times 10^-19 C

F = 1.6 times 10^-19 (v times B)

5.9 times 10^-17