# The magnetic force (FM) on a particle in a magnetic field is found by FM=I*B, I is the charge multiplied by the velocity of a charged particle and B is the strength of the magnetic field, in...

The magnetic force (FM) on a particle in a magnetic field is found by FM=I*B, I is the charge multiplied by the velocity of a charged particle and

B is the strength of the magnetic field, in Tesla (T). An electron accelerates from rest to the right, in a horizintally directed electric field. The electron then leaves the electric field at a speed of 4.0 x 10^6, entering a magnetic field of magnitude 0.20 T (Tesla) directed into the screen. Calculate the magnitude and direction of the magnetic force on the electron. (Charge on an electron: q=1.6 x 10^-19)

*print*Print*list*Cite

### 1 Answer

The magnetic force is given by FM = I*B, where I is the charge multiplied by the velocity of the charged particle.

I = 4.0*10^6*1.6*10^-19

The magnetic field is equal to 0.2 Tesla directed in a direction into the screen

The magnitude of the magnetic force is 4.0*10^6*1.6*10^-19*0.2

=> 1.28*10^-13

The formula for the magnetic field is given as a dot product, FM = I*B. There is no information provided about the interaction between the velocity and the direction of the magnetic field that would determine the direction of the magnetic force. So we can take the the direction of the magnetic field would be the same as that of velocity.

**The magnitude of the magnetic force is 1.28*10^-13 and its direction is in the horizontal direction towards the right.**