Using the formulas for the force on a particle of charge, e, moving perpendicula

Question

Using the formulas for the force on a particle of charge, e, moving perpendicular to a magnetic field, B, and the centripetal force on any object moving in a circle, show that |r| = mv / (eB). Using conservation of energy, find the velocity of a particle with mass m and charge e after it has been accelerated through a potential different of V. Using your answers from parts (1) and (2), show that the charge to mass } ratio (e/m) for moving particle described is: e/m = 2V/(r^2B^2) In this lab we will be using a magnetic from a particular apparatus referred to as a Helmhotz coil. This is a set of two parallel sets of coils of wire of radius R and N coils. The two coils are placed a distance R apart. Using the formula for the magnetic field at the center of coil of wire, show that the magnetic field at the a point, p, midway between the center of the two coils is equal to B=(4/5)^3/2/ mu_0 NI/R Using the formula above, what is the magnetic field/ampere for a set of coils with 130 turns and a radius of 0.152m?

Explanation / Answer

1)
magnetic field must provide necessary cenripetal force for moving in circle
magnetic force = e*v*B
centripetal forec = m*v^2/r

Balancing both:
e*v*B=m*v^2/r
e*B=m*v/r
r=m*v/e*B

2)
After it has been accelerated through potential V, its potential energy will change by e*V
Let its kinetic energy change be 0.5*m*vf^2
Use conservation of energy,
change in K.E= Change in P.E
0.5*m*v^2 = e*V
v^2 = 2*e*V/m
v = sqrt (2*e*V/m)

3.
from part 1:
r=m*v/e*B
v = r*e*B/m
from part 2:
v = sqrt (2*e*V/m)

equatiing both v
r*e*B/m = sqrt (2*e*V/m)
r^2*B^2*(e/m)^2 =2*V*(e/m)
r^2*B^2*(e/m) =2*V
e/m =2*V/(r^2*B^2)

Rest 2 are not related to above and hence will form diffrent question

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