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

a) period = 0.364 ns

b) pitch
`=1.44 * 10^(-3)`

c) radius
`=1.63 * 10^(-3) m`

Step-by-step explanation:

Given data:

mass of the positron is
`m = 9.1 * 10^(-31) kg=`

charge is
`q = 1.6 * 10^(-19)C`

strength of the magnetic field is B = 0.0980 T

kinetic energy is
`K=2.30 keV = 2.30 * 10^3 * 1.6 * 10^(-19)J`

`0.5mv^2 = 3.68 * 10^(-16) J`

`0.5 * 9.1 * 10^(-31) * v^2 = 3.68 x 10^(-16)`

`v= 2.843 * 10^7m/s`

a) period is
`T=(2πm)/(qB)`

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

=0.364 ns

b)pitch is
`p = vcos\theta ((2πm)/(qB))`

`=2.843 * 10^7 * cos82(0.3646 * 10^(-9))`

`=1.44 * 10^(-3)`

c) radius is
`r = (mvsin\theta)/(qB)`

`=((9.1 * 10^(-31))* 2.843 * 10^7* sin 82)/(1.6 * 10^(-19)* 0.0980)`

`=1.63 * 10^(-3) m`

Answer 2

a)0.364 ns

b)1.43 x 10^-3

c)163.21 x 10^-5 m

Step-by-step 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