Question

A positron with kinetic energy 2.30 keV is projected into a uniform magnetic field of magnitude 0.0980 T, with its velocity vector making an angle of 82.0° with the field. Find (a) the period, (b) the pitch p, and (c) the radius r of its helical path.

Answer 1

Answer:

a) period = 0.364 ns

b) pitch  $=1.44 \times 10^{-3}$

c) radius $=1.63 \times 10^{-3} m$

Explanation:

Given data:

mass of the positron is $m = 9.1 \times 10^{-31} kg=$

charge is $q = 1.6 \times 10^{-19}C$

strength of the magnetic field is B = 0.0980 T

kinetic energy is $K=2.30 keV = 2.30 \times 10^3 \times 1.6 \times 10^{-19}J$

$0.5mv^2 = 3.68 \times 10^{-16} J$

$0.5 \times 9.1 \times 10^{-31} \times v^2 = 3.68 x 10^{-16}$

$v= 2.843 \times 10^7m/s$

a) period is $T=\frac{2πm}{qB}$

                         $=\frac{2π(9.1 \times 10^{-31})}{1.6 \times 10{-19}\times 0.0980}$

                         =0.364 ns

b)pitch is $p = vcos\theta \frac{(2πm}{qB})$

                  $=2.843 \times 10^7 \times cos82(0.3646 \times 10^{-9})$

$=1.44 \times 10^{-3}$

c) radius is $r = \frac{mvsin\theta}{qB}$

                  $=\frac{(9.1 \times 10^{-31})\times 2.843 \times 10^7\times sin 82}{1.6 \times 10^{-19}\times 0.0980}$

                  $=1.63 \times 10^{-3} m$

Answer 2

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