Question

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electron is 4.90 10-19 J, find the speed of the electron and the radius of the circular path.

Answer 1

The speed of the electron and the radius of the circular path are
`10.38*10^(5)\ m/s` and
`2.15*10^(-5)\ m`.

Step-by-step explanation:

Given that,

Magnetic field = 0.275 T

Kinetic energy
`K.E= 4.90*10^(-19)\ J`

We need to calculate the speed of the electron

Using kinetic energy

`K.E=(1)/(2)mv^2`

`v=\sqrt{(2K.E)/(m)}`

Where, m = mass of electron

K.E = kinetic energy

Put the value into the formula

`v= \sqrt{(2*4.90*10^(-19))/(9.1*10^(-31))}`

`v=10.38*10^(5)\ m/s`

We need to calculate the radius of the circular path

Using equation of force on charge in magnetic field

`F = qvB\sin\theta`....(I)

Using centripetal force

`F=(mv^2)/(r)`....(II)

From equation (I) and (II)

`(mv^2)/(r)=qvB\sin\theta`

`r=(mv^2)/(qvB\sin\theta)`

Put the value into the formula

`r=(9.1*10^(-31)*10.38*10^(5))/(1.6*10^(-19)*0.275\sin90^(\circ))`

`r=(9.1*10^(-31)*10.38*10^(5))/(1.6*10^(-19)*0.275)`

`r=2.15*10^(-5)\ m`

Hence, The speed of the electron and the radius of the circular path are
`10.38*10^(5)\ m/s` and
`2.15*10^(-5)\ m`.

Answer 2

Radius,
`r=2.14* 10^(-5)\ m`

Step-by-step explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron,
`E=4.9* 10^(-19)\ J`

Kinetic energy is given by :

`E=(1)/(2)mv^2`

`v=\sqrt{(2E)/(m)}`

`v=\sqrt{(2* 4.9* 10^(-19))/(9.1* 10^(-31))}`

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

`qvB\ sin90=(mv^2)/(r)`

`r=(mv)/(qB)`

`r=(9.1* 10^(-31)* 1037749.04)/(1.6* 10^(-19)* 0.275 )`

`r=2.14* 10^(-5)\ m`

So, the radius of the circular path is
`2.14* 10^(-5)\ m`. Hence, this is the required solution.