Answer:
a)0.364 ns
b)1.43 x 10^-3
c)163.21 x 10^-5 m
Explanation:
mass of the positron is m=9.1 x 10^-31 kg
charge is q=1.6 x 10^-19 C
strength of the magnetic field is B=0.0980 T
kinetic energy is K=2.30 keV=2.30 x 10^3 x 1.6 x 10^-19 J
0.5mv^2=6.68 x 10^-16 J
0.5 x 9.1 x 10^-31 x v^2 =3.68 x 10^-16
v=2.84 x 10^7 m/s
a)the period is T=2πm/qB
=2π(9.1 x 10^-31 )/1.6 x 10^-19x 0.0980
=0.364 ns
b) pitch is p=vcosθ(2πm/qB)
=2.84 x 10^7x cos82(0.364 x 10^-9)
=1.43 x 10^-3
c) radius is r=mvsinθ/qB
=(9.1 x 10^-31 )2.84 x 10^7x sin82/1.6 x 10^-19x 0.098
=163.21 x 10^-5 m
Answer:
a) period = 0.364 ns
b) pitch [tex]=1.44 \times 10^{-3}[/tex]
c) radius [tex] =1.63 \times 10^{-3} m[/tex]
Explanation:
Given data:
mass of the positron is [tex]m = 9.1 \times 10^{-31} kg=[/tex]
charge is [tex]q = 1.6 \times 10^{-19}C[/tex]
strength of the magnetic field is B = 0.0980 T
kinetic energy is [tex]K=2.30 keV = 2.30 \times 10^3 \times 1.6 \times 10^{-19}J[/tex]
[tex]0.5mv^2 = 3.68 \times 10^{-16} J[/tex]
[tex]0.5 \times 9.1 \times 10^{-31} \times v^2 = 3.68 x 10^{-16}[/tex]
[tex]v= 2.843 \times 10^7m/s[/tex]
a) period is [tex]T=\frac{2πm}{qB}[/tex]
[tex]=\frac{2π(9.1 \times 10^{-31})}{1.6 \times 10{-19}\times 0.0980}[/tex]
=0.364 ns
b)pitch is [tex]p = vcos\theta \frac{(2πm}{qB})[/tex]
[tex]=2.843 \times 10^7 \times cos82(0.3646 \times 10^{-9})[/tex]
[tex]=1.44 \times 10^{-3}[/tex]
c) radius is [tex]r = \frac{mvsin\theta}{qB}[/tex]
[tex]=\frac{(9.1 \times 10^{-31})\times 2.843 \times 10^7\times sin 82}{1.6 \times 10^{-19}\times 0.0980}[/tex]
[tex] =1.63 \times 10^{-3} m[/tex]
