Use the following equations to derive the equation for e/m = 2V(5/4)3a2/(N mu s

Question

Use the following equations to derive the equation for e/m = 2V(5/4)3a2/(N mu s Ir)2 Fm rightarrow = qv rightarrow times B rightarrow where Fm rightarrow = the magnetic force acting on a charged q = charge on the particle, v rightarrow = velocity, B rightarrow = magnetic field eV = 1/2 mv2 Fe = mv2/r and B = [N mu s]I/(5/4)3/2 sigma for the Helmholtz Colls. where a = the radius of the Helmholtz coils (0.15 m) N = the number of the turns on each Helmholtz coil (130 turns) mu s = permeability constant = 4 times 10-7 = T.m/A I = the current through the Helmholtz coils r = the radius of the electron beam path

Explanation / Answer

First start with Lorentz formula:
F = qvBsin(theta) in this case theta = 90 (sin(90)=1) and q (charge) is actually e Thus:
F = evB

Now, we also know the centripetal force:
F = (mv^2)/(r)
Substitute Fc to Lorentz formula:
(mv^2)/(r) = evB
or:
(mv^2) = evBr

Formula 1

mv = eBr

But we do not know velocity so...

Yet again we know what that E = 1/2mv^2 where E = qV or E= eV (charge x volts)

So:

eV = 1/2 mv^2

v = sqrt((2eV)/m) **where v = velocity and V = volts

Substitute velocity into Formula 1 and we get:

mv = eBr

m*sqrt((2eV)/m) = eBr

Square both sides to get rid of the square root:

m^2 * ((2eV)/m) = e^2 * B^2 * r^2 **(e^2 * B^2 * r^2 can also equal (eBr)^2 )

The masses cancel on one side and take the left e to the right to cancel e and we get:

m2V = e * B^2 * r^2

Take m to be on the same side as e and B^2 * r^2 to the otherside and we get:

(e/m) = ((2V)/(B^2 * r^2 ))

We know B = ((Nu0)I)/ ((5/4)^(3/2)*a)) now substiture for B

(e/m) = ((2V)/(((Nu0)I)/ ((5/4)^(3/2)*a))^2 * r^2 ))

Simplify and we get:

(e/m) =  ((2V*(5/4)^3 * a^2)/((Nu0I r)^2 ))

Which is what you wanted to prove.

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