Question

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters perpendicularly to the field. Under these conditions, the electron undergoes circular motion. Find the radius r of the electron's path and the frequency f of the motion.

Answer 1

(a) 422.2 x 10⁻⁷m

(b) 4.6 x 10⁵ Hz

Step-by-step explanation:

The magnetic force acting perpendicular to the velocity of the electron will cause a circular motion so that the centripetal force,
`F_(c)`, of the electron is equal to the Lorentz's force,
`F_(l)`. i.e

`F_(c)` =
`F_(l)` ---------------(i)

Where;

`F_(c)` =
`(mv^2)/(r)`

`F_(l)` = qvB

Equation (i) then becomes;

`(mv^2)/(r)` = 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 =
`(mv^2)/(qvB)`

r =
`(mv)/(qB)` -----------(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 =
`(9.1*10^(-31) * 121)/(1.6*10^(-19)*1.63*10^(-5))`

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 =
`(v)/(2\pi r)`

Substitute the values of v and r into the equation as follows;

f =
`(121)/(2\pi (422.2*10^(-7)))`

Take
`\pi` = 3.142

f =
`(121)/(2(3.142) (422.2*10^(-7)))`

f = 4.6 x 10⁵ Hz

Therefore, the frequency of the motion is 4.6 x 10⁵ Hz

Answer 2

i. The radius 'r' of the electron's path is 4.23 ×
`10^(-5)` m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Step-by-step 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 =
`(mv)/(qB)`

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 ×
`10^(-5)`T, v = 121 m/s, Θ =
`90^(0)` (since it enters perpendicularly to the field), q = e = 1.6 ×
`10^(-19)`C and m = 9.11 ×
`10^(-31)`Kg.

Thus,

r =
`(mv)/(qB)` ÷ sinΘ

But, sinΘ = sin
`90^(0)` = 1.

So that;

r =
`(mv)/(qB)`

= (9.11 ×
`10^(-31)` × 121) ÷ (1.6 ×
`10^(-19)` × 1.63 ×
`10^(-5)`)

= 1.10231 ×
`10^(-28)` ÷ 2.608 ×
`10^(-24)`

= 4.2266 ×
`10^(-5)`

= 4.23 ×
`10^(-5)` m

The radius 'r' of the electron's path is 4.23 ×
`10^(-5)` m.

B. The frequency 'f' of the motion is called cyclotron frequency;

f =
`(qB)/(2\pi m)`

= (1.6 ×
`10^(-19)` × 1.63 ×
`10^(-5)`) ÷ (2 ×
`(22)/(7)` × 9.11 ×
`10^(-31)`)

= 2.608 ×
`10^(-24)` ÷ 5.7263 ×
`10^(-30)`

= 455442.4323

f = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.