In a mass spectrometer, a beam of protons enters a magnetic field. Some protons

Question

In a mass spectrometer, a beam of protons enters a magnetic field. Some protons make exactly a one-quarter circular arc of radius 0.45 m .

Part A) If the field is always perpendicular to the proton's velocity, what is the field's magnitude if exiting protons have a kinetic energy of 11 keV ?

Part B) How long does it take the proton to complete the quarter circle? Express your answer using two significant figures.

Part C) Find the net force (magnitude) on a proton while it is in the field. Express your answer using two significant figures.

Explanation / Answer

Given,

R = 0.45 m

a) KE = 11 keV = 11000 eV = 1.76 x 10-15 J

We need to determine the maggnitude of magnetic field B.

We know that, KE = 1/2 m v2

1/2 m v2 =  1.76 x 10-15 J => v2 = 2 x  1.76 x 10-15 J / m

v = sqrt ( 2 x  1.76 x 10-15 J /  9.1 x 10-31 J ) = 6.22 x 107 m/s

We know that, in mass spectrometer, due to the presence of magnetic field, the moving particle follows a circular path and the centripital force gets balanced by the magnetic force.

m v2/R = q v B

B = m v / q R = 9.1 x 10-31 x 6.22 x 107 / 1.6 x 10-19 x 0.45 = 7.86 x 10-4 Tesla

Hence, B =  7.86 x 10-4 Tesla

B) Let this time be "t"

We know that, speed = distance/time => time = distance/speed

distance = 1/4 (circumference of the circle) = 1/4 x 2 pi R = pi R / 2 = 3.14 x 0.45 / 2 = 0.7065 m

time = t = 0.7065 /  6.22 x 107 = 0.114 x 10-7 sec = 1.14 x 10-8 sec

Hence, t = 1.14 x 10-8 sec.

C) The net force on the proton will be zero when it is in the field. As we discussed earliear that the magnetic force ( Fm = q v B) balances the centripital force (Fc = mv2/R) and there is not other forces acting on the proton. So net force on the proton will be zero.

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