Answer:
i. The radius 'r' of the electron's path is 4.23 × [tex]10^{-5}[/tex] m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r = [tex]\frac{mv}{qB}[/tex]
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 × [tex]10^{-5}[/tex]T, v = 121 m/s, Θ = [tex]90^{0}[/tex] (since it enters perpendicularly to the field), q = e = 1.6 × [tex]10^{-19}[/tex]C and m = 9.11 × [tex]10^{-31}[/tex]Kg.
Thus,
r = [tex]\frac{mv}{qB}[/tex] ÷ sinΘ
But, sinΘ = sin [tex]90^{0}[/tex] = 1.
So that;
r = [tex]\frac{mv}{qB}[/tex]
= (9.11 × [tex]10^{-31}[/tex] × 121) ÷ (1.6 × [tex]10^{-19}[/tex] × 1.63 × [tex]10^{-5}[/tex])
= 1.10231 × [tex]10^{-28}[/tex] ÷ 2.608 × [tex]10^{-24}[/tex]
= 4.2266 × [tex]10^{-5}[/tex]
= 4.23 × [tex]10^{-5}[/tex] m
The radius 'r' of the electron's path is 4.23 × [tex]10^{-5}[/tex] m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f = [tex]\frac{qB}{2\pi m}[/tex]
= (1.6 × [tex]10^{-19}[/tex] × 1.63 × [tex]10^{-5}[/tex]) ÷ (2 ×[tex]\frac{22}{7}[/tex] × 9.11 × [tex]10^{-31}[/tex])
= 2.608 × [tex]10^{-24}[/tex] ÷ 5.7263 × [tex]10^{-30}[/tex]
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.
(a) 422.2 x 10⁻⁷m
(b) 4.6 x 10⁵ Hz
The magnetic force acting perpendicular to the velocity of the electron will cause a circular motion so that the centripetal force, [tex]F_{c}[/tex], of the electron is equal to the Lorentz's force, [tex]F_{l}[/tex]. i.e
[tex]F_{c}[/tex] = [tex]F_{l}[/tex] ---------------(i)
Where;
[tex]F_{c}[/tex] = [tex]\frac{mv^2}{r}[/tex]
[tex]F_{l}[/tex] = qvB
Equation (i) then becomes;
[tex]\frac{mv^2}{r}[/tex] = qvB ------------------(ii)
Where;
m = mass of the electron
v = linear velocity of the electron
r = radius of the electron's path
q = charge of the electron
B = magnetic field.
Make r subject of the formula in equation (ii)
r = [tex]\frac{mv^2}{qvB}[/tex]
r = [tex]\frac{mv}{qB}[/tex] -----------(iii)
From the question;
v = 121m/s
B = 1.63 x 10⁻⁵ T
q = 1.6 x 10⁻¹⁹ C (known constant)
m = 9.1 x 10⁻³¹kg
Substitute these values into equation (iii) as follows;
r = [tex]\frac{9.1*10^{-31} * 121}{1.6*10^{-19}*1.63*10^{-5}}[/tex]
r = 422.2 x 10⁻⁷m
Therefore, the radius of the electron's path is 422.2 x 10⁻⁷m
(ii) The frequency, f, of the motion which is also called the cyclotron frequency - the number of cycles the electron completes around the path every second - is given by;
f = [tex]\frac{v}{2\pi r}[/tex]
Substitute the values of v and r into the equation as follows;
f = [tex]\frac{121}{2\pi (422.2*10^{-7})}[/tex]
Take [tex]\pi[/tex] = 3.142
f = [tex]\frac{121}{2(3.142) (422.2*10^{-7})}[/tex]
f = 4.6 x 10⁵ Hz
Therefore, the frequency of the motion is 4.6 x 10⁵ Hz
