Question

Example 4.3 What is the radius of the path of an electron (mass
9x10 ^-31 kg and charge 1.6 x 10^-19 C) moving at a speed of 3 x10^7 m/s in
a magnetic field of 6 X 10^-4 T perpendicular to it? What is its
frequency? Calculate its energy in keV. (1 eV = 1.6x10^-19 J).​

Answer 1

Answer:

The force experienced by the electron will be equal to the centripetal force required to keep the electron in its circular path.

So,

evB = mv2/r

=> r = mv/eB

=> r = 0.28125 m

Thus, the circumference of the circular path is = 2 × π × r = 1.76625 m

Time period of revolution = 1.76625/v = 5.8875 × 10-8 s

So, frequency = 1/(5.8875 × 10-8) = 1.7 × 107 Hz

Energy of the electron is = ½ mv2 = 4.05 × 10-16 J

So, the energy in eV is = (4.05 × 10-16)/(1.6 × 10-19) = 2531.25 eV

Hope it helps you

Answer 2

Answer:

radius=28cm,frequency=17MHz and energy=2.53keV

Explanation:

Here, m=9×10−31kg,

q=1⋅6×10−19C, v=3×107ms−1,

B=6×10−4T

r=mvqB=(9×10−31)×(3×107)(1⋅6×10−19)(6×10−4)=0⋅28m

v=v2πr=Bq2πm=(6×10−4)×(1⋅6×10−19)2×(22/7)×9×10−31

=1⋅7×107Hz

Ek=12mv2=12×(9×10−31)×(3×107)2J

=40⋅5×10−17J=40⋅5×10−171⋅6×10−16keV

=2⋅53keV