A proton with a velocity V = (2.00 m / s) i - (4.00 m /...

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)

Solutions

Expert Solution

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.

genius_generous answered 1 year ago

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