A uniform magnetic field of magnitude B=1.2 teslas is directed toward the bottom of the page in the -y direction. At time t=0, a proton p in the field is moving in the plane of the page with a speed v=4 x 10^7 meters per second in a direction 30 degrees above the +x axis.
a.) Calculate the magnetic force on the proton at t=0
b.) With reference to the coordinate system shown above on the right (see attachment), state the direction of the force on the proton at t=0
c.) How much work will the magnetic field do on the proton during the interval from t=0 to t=0.5 seconds?
b) When a particle of charge q moves in a magnetic field B with a velocity v, the magnetic force on the particle is given by
F = qvB
The force, velocity and the magnetic field are all perpendicular to each other.
In the case above, the velocity of the proton is not perpendicular to the magnetic field. In this case we have to resolve the velocity and use the component that is perpendicular to the field. In this case the perpendicular component is the one along the +x axis.
q = +1 ...
This solution of 391 words contains both calculations and justifications in determining the magnetic force, direction and magnetic field of the proton. All steps are shown clearly with full workings.