FL = B q v
- FL magneticforce in N
- B magnetic induction in T
- q charge particle in C
- v speed particle in m s-1
An electron enters at A in a uniform magnetic field with a speed of 4.0 x 107 m.s-1
The magnetic field has a magnitude of 3.0 x 10-3 T and the direction is into the paper
The electrons move perpendicular to the magnetic field.
On the electron acts a magnetic force :
FL = B q v
FL = (3.0 x 10-3)( 1.6 x 10-19)( 4.0 x 107) = 1.92 x 10-14 N
The Left-Hand-Rule determines the direction of the magnetic force . (For the description of this rule : see example 1 in the previous paragraph)
The electron goes at A from the left side to the right side.
The direction of the current I is opposite to the direction of the velocity of the electron.
So the direction of I is from right to left.
The direction of the magnetic force is downward.
F is perpendicular to the velocity v.
The electron is in a uniform circular motion.
Calculate the radius of the path
FL is the centripetal force
FL = Fc
B q v = (m v2)/r
r = (m v)/(B q)
r = ((9.1 x 10-31 )( 4.0 x 107)) / ((3.0 x 10-3)( 1.6 x 10-19 )) = 7.6 x 10-2 m