In: Physics
A proton with a velocity V = (2.00 m / s) i - (4.00 m / s) j - (1.00 m / s) k, a B = (1.00 T) i + (2.00 T) j- (1.00 T) k it moves within the magnetic field. What is the magnitude of the magnetic force (Fe) acting on the particle? (Qproton = 1.6x10-19 C)
When a charged particle enter into a magnetic field it experiences a force called lorentz force. And force can be written as
Here q is the charge of the particle, V is the velocity of the particle and B is the magnetic field . First we take the cross product between V and B. And then we multiply it with the charge q and then by taking the magnitude of the vector quantity we get the magnitude of the force. The detailed solution is shown below as an image .
So the magnitude of magnetic force acting on the particle is 16.08 ×10^-19 N.