Question

​ An electron is accelerated from rest through 2300 V and then enters a region of uniform magnetic field with magnitude 1.25 T. What is the magnitude of the maximum magnetic force acting on the electron?

Answer 1

`F_(maximum) =5.68*10^(-12) \ \ N`

Step-by-step explanation:

Electron acquire through kinetic energy is given as:

`K = (1)/(2)mv^2\\ v^2 = (2 \ K)/(m)\\v = \sqrt{(2 \ K)/(m)}`

Given that :

Electron energy expanded = 2300 V

= 2300 eV

`= 2300 *1.6*10^(-19) \ J`

=
`3.68*10^(-16) \ J`

Velocity acquired
`v = \sqrt{(2 K)/(m)}`

`v = \sqrt{(2 *3.66*10^(-16) )/(9.11*10^(-31))}`

`v = 28423641.62`

`v = 2.84 *10^7 \ m/s`

Force exerted by the magnetic field

`F_B = Bvq sin \theta`

where θ = 90° and sin θ = sin 90 ° = 1

∴

`F_(maximum) = 1.6*10^(-19) * 2.84*10^7 *1.25`

`F_(maximum) =5.68*10^(-12) \ \ N`

Answer 2

Step-by-step explanation:

Given that,

The electron accelerated from rest through 2300V, then the Potential difference is 2300V

V = 2300V.

Magnetic field of 1.25T

B= 1.25T

Mass of electron

m = 9.11 ×10^-31kg

Charge of an electron

q = 1.602 ×10^-19C

Maximum magnetic force?

Maximum magnetic force can be determined using

Fmax = qvB

We don't know the velocity(v) but it can be determine using

Work done by potential = Kinectic energy

qV = ½mv²

Rearranging

v = √(2qV/m)

v = √(2×1.602×10^-19 × 2300/9.11×10^-31)

v = 2.84 × 10^7 m/s

Then,

Maximum Magnetic force

Fmax = qvB

Fmax=1.602×10^-19×2.84×10^7×1.25

Fmax = 5.7 × 10^-12 N

The magnitude of the maximum magnetic force is 5.7 × 10^-12 N