In an experiment with cosmic rays, a vertical beam of particles that have charge

Question

In an experiment with cosmic rays, a vertical beam of particles that have charge of magnitude 3e and mass 12 times the proton mass enters a uniform horizontal magnetic field of 0.110 T and is bent in a semicircle of diameter
d = 93.0 cm, as shown in the figure below.

(a) Find the speed of the particles and the sign of their charge.

(b) Is it reasonable to ignore the gravity force on the particles?

Yes No    

(c) How does the speed of the particles as they enter the field compare to their speed as they exit the field?

They exit with greater speed. They exit with less speed.     They exit with the same speed.

Explanation / Answer

q = charge = 3 e = 3 x 1.6 x 10-19 C

m = mass = 12 x 1.67 x 10-27 kg

B = magnetic field = 0.11 T

d = diameter = 93 cm

r = radius = d/2 = 93/2 = 46.5 cm = 0.465 m

a)

using the formula

r = mv/(qB)

0.465 = (12 x 1.67 x 10-27 ) v / (3 x 1.6 x 10-19 x 0.11)

v = 1.225 x 106 m/s

magnetic field experienced is in right direction as the particle enters the magnetic field in down direction

hence the charge is positive.

b)

the mass of particle is very small hence we can ignore the force of gravity.

c)

the same speed as the force only act to change the direction of particles path and does not change its speed.

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