What uniform magnetic field, applied perpendicular to a beam of electrons moving

Question

What uniform magnetic field, applied perpendicular to a beam of electrons moving at 1.30 Times 106 m/s, is required to make the electrons travel in a circular are of radius 0.350 m? An electron is accelerated from rest by a potential difference mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field. Find the frequency of revolution of an electron with an energy of 100 eV in a uniform magnetic field of magnitude 35.0 muT. Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field. In fig. 28-38, a charged particle moves into a region of uniform magnetic field , goes through half a circle, and then exits that reglion. The particle is either a proton or an electron (you must decide which). It spends 130 ns in the region. (a) What is the magnitude of ? (b) If the particle is sent back through the magnetic field (along the same initial path) but with 2.00 times its previous kinetic energy, how much time does it spend in the field during this trip? A mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass 3.92 Times 10-25kg and charge 3.20 Times 10-19 C from related species. The ions are accelerated through a potential difference of 100 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.00 m. After traveling through 180 degree and passing through a slit of width 1.00 mm and height 1.00 cm, they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine

Explanation / Answer

Part A)

Since the force pushes the charged particle to the left, the right hand rule tells us that the particle must have a positive charge. Thus the particle is a proton.

The distance is half a circle, which is (pi)(r)

We can apply d = vt and theus (pi)(r) = (v)t or r = vt/pi

The formula for the radius of a charged paticle in a magnetic field is

r = mv/qB

By substitution...

vt/pi = mv/qB (v cancels)

(130 X 10^-9)/(pi) = (1.67 X 10^-27)/(1.6 X 10^-19)(B)

B = .252 T

Part B)

Since KE = .5mv^2, if KE is doubled, the velocity is increased by the square root of 2 (or 1.414)

However, looking at the above derived formula, we can see that velocity does not factor in since it cancelled out, so it would still spend 130 ns in the field

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