Question

An electron is accelerated through 1.95 103 V from rest and then enters a uniform 1.50-T magnetic field. (a) What is the maximum magnitude of the magnetic force this particle can experience?

Answer 1

6.3 x 10⁻¹²N

Step-by-step explanation:

As stated by Lorentz Force law, the magnitude of a magnetic force, F, can be expressed in terms of a fixed amount of charge, q, which is moving at a constant velocity, v, in a uniform magnetic field, B, as follows;

F = qvB sin θ ------------(i)

Where;

θ = angle between the velocity and the magnetic field vectors

When the electron passes through a potential difference, V, it is made to accelerate as it gains some potential energy (
`P_(E)`) which is then converted to kinetic energy (
`K_(E)`) as it moves. i.e

`P_(E)` =
`K_(E)` ----------------(ii)

But;

`P_(E)` = qV

And;

`K_(E)` =
`(1)/(2)` x m x v²

Therefore substitute these into equation (ii) as follows;

qV =
`(1)/(2)` x m x v²

Make v subject of the formula;

2qV = mv²

v² =
`(2qV)/(m)`

v =
`\sqrt{(2qV)/(m) }` ---------------(iii)

From the question;

q = 1.6 x 10⁻¹⁹C (charge on an electron)

V = 1.95 x 10³V

m = 9.1 x 10⁻³¹kg

Substitute these values into equation (iii) as follows;

v =
`\sqrt{(2*1.6*10^(-19) * 1.95*10^(3))/(9.1*10^(-31)) }`

v =
`\sqrt{(6.24*10^(-16))/(9.1*10^(-31)) }`

v = 2.63 x 10⁷m/s

Now, from equation (i), the magnitude of the magnetic force will be maximum when the angle between the velocity and the magnetic field is 90°. i.e when θ = 90°

Substitute the values of θ, q v and B = 1.50T into equation (i) as follows;

F = qvB sin θ

F = 1.6 x 10⁻¹⁹ x 2.63 x 10⁷ x 1.50 x sin 90°

F = 6.3 x 10⁻¹²N

Therefore the maximum magnitude of the magnetic force this particle can experience is 6.3 x 10⁻¹²N

Answer 2

Answer: 6.29*10^-12 N

Step-by-step explanation:

given,

Potential difference of the electron, v = 1950 V

Magnetic field of the electron, B = 1.50 T

If the electron is accelerated through 19500 V from rest its Potential Energy has to be

converted to Kinetic Energy. This allows us solve for the velocity.

PE = Vq

PE = 1950 * 1.6*10^-19

PE = 3.12*10^-16 J

Also, PE = 1/2mv²

3.12*10^-16 = 1/2mv²

v = 2.62*10^7 m/s

to get F(max), we use,

F(max) = qvB

F(max) = 1.6*10^-19 * 2.62*10^7 * 1.5

F(max) = 6.29*10^-12 N