2. -4 points My Notes Ask Your Teacher An electron is accelerated from rest by a

Question

2. -4 points My Notes Ask Your Teacher An electron is accelerated from rest by a potential difference of 350 V. It then enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular to the field. (a) Calculate the speed of the electron as it enters the magnetic field. (b) Calculate the radius of its path in the magnetic field. (c) Suppose another elementary partical, a muon (mass 206.8 x mass of electron and same charge) is accelerated from rest in the same apparatus. Calculate the speed of the muon as it enters the magnetic field. (d) Calculate the radius of the path of the muon in the magnetic field.

Explanation / Answer

Potential energy drop = qV
=
Kinetic energy gained = 1/2 mv^2

Solve for the speed of the electron:
v = sqrt (2qV/m)

so v=sqrt(2*1.6*10^-19*350/9.1*10^-31)

=sqrt(123.07*10^12)

=11.09*10^6

v=1.109*10^5 m/sec--answer to a

b)

Centrifugal force = mv^2 / r

Magnetic force = qvB

both will be equal

mv^2/r=qvB

v=qBr/m

r=mv / qB

r=9.1*10^-31*1.109*10^5/1.6*10^-19*200*10^-3

r=0.031*10^-4

r=0.0031mm--answer to b

c)here mass of moun will be

mass=206.8*9.1*10^-31

=1881.88*10^-31

=1.8*10^-28Kg

so v = sqrt (2qV/m)

=sqrt(2*1.6*10^-19*350/1.8*10^-28)

=sqrt(622.3*10^9)

=sqrt(62.23*10^8)

v=7.89*10^4 m/sec--answer to c

d)r=mv/qB

r=1.8*10^-28*7.89*10^4/1.6*10^-19*200*10^-3

r=0.045*10^-2

r=0.45*10^-3

r=0.45mm--answer to d

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