A mass spectrometer is used to measure the mass of charged particles. Initially

Question

A mass spectrometer is used to measure the mass of charged particles. Initially at rest, a beam of charged particles, each with q= +1.6 x 10^-19 C and m = 1.67 x 10^27 kg, is accelerated through a small aperture across a charged capacitor (V=1000C), following which the particle then enters a magnetic field.

A. Determine the magnetic force vector exerted on the charge, in component form, as it enters the field.

B. The particle completes the semicircular path within the B field and its displacement is measured using the sensor. How far from the entry point does the particle reach the sensor?

C. A second particle, same charge, is then measured with the device. In order for this particle to reach the sensor (same position as above), the B field is increased to 0.62T. What is the mass of the particle?

D. An important experiment in the development of electromagnetism was the measurement of the mass to charge ratio of the electron, m/e. Show (derive) that this ratio is equal to (B^2*r^2)/2V.

3. Amass spectrometer is used to measure the mass of charged particles. Initially at rest, a beam of charged particles, each with q = +1.6 x 101°C and m = 1.67 x 10-27 kg, is accelerated through a small aperture across a charged capacitor (V = 1000 V), following which the particle then enters a magnetic field V-1000V x (out of page) Sensor Determine the magnetic force vector exerted on the charge, in component form, as it enters the a. field. b. The particle completes the semicircular path within the B field and its displacement is measured using the sensor. How far from the entry point does the particle reach the sensor? C. A second particle, same charge, is then measured with the device. In order for this particle to reach the sensor (same position as above), the B field is increased to 0.62T. What is the mass of the particle? An imp ortant experiment in the development of electromagnetism was the measurement of the d. that this ratio is equal to B'r'/2V mass to charge ratio of the electron, me/e. Show (derive) that this rati

Explanation / Answer

in the electric field

W = V*q = (1/2)*m*v^2

speed v = sqrt(2*V*q/m) = sqrt(2*1000*1.6*10^-19/(1.67*10^-27)) = 437740.5 m/s

magnetic force F = q*( v x B )

q = 1.6*10^-19 C

velocity v = 437740.5 j m/s

magntic field B = 0.1 i

Fb = 1.6*10^-19*( 437740.5j x 0.1 i)

Fb = - 7*10^-15 k N

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(b)

In the magneitc field Fb = Fc

Fb = m*v^2/r

radius r = mv/Fb

distance d = 2r = 2*mv^2/Fb

d = 2*1.67*10^-27*437740.5^2/(7*10^-15)

d = 0.091 m

==============================================

part( c )

Fb = Fc

q*v*B = mv^2/r

r = mv/(qB)

r1 = m1*v1/(q1*B1)

r2 = m2*v2/(q2*B2)

given

r2 = r1

q2 = q1

B1 = 0.1

B2 = 0.62

m1*V1/B1 = m2*v2/B2

m2 = m1*(v1/v2)*(B2/B1)

speed v1 = sqrt(2*Vq/m1)

speed v2 = sqrt(2*V*q2/m2)

v1/v2 = sqrt(m1/m2)

therefore

m2 = m1*sqrt(m1/m2)*(B2/B1)

m2 = 1.67*10^-27*sqrt(1.67*10^-27/m2)*(0.62/0.1)

mass of second particle

m2 = 5.64*10^-27 kg <<<<<<--------------ANSWER

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In the electric field

work = dK

V*q = (1/2)*m*v^2

v^2 = 2*V*q/m

magnetic force Fb = centripetal force

q*v*B = m*v^2/r

m*v = q*B*r

m^2*v^2 = q^2*B^2*r^2

m^2*2V*q/m = q^2*B^2*r^2

m*2*V = q*B^2*r^2

m/q = B^2*r^2/2V

charge q = e

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

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