Question

An electron with kinetic energy 1.40 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 30.0 cm. Find (a) the speed of the electron, (No Response) m/s (b) the magnetic field, (No Response) T (c) the frequency of circling, and (No Response) Hz (d) the period of the motion.

Answer 1

(a)
`v=22.177* 10^(12)\ m.s^(-1)`

(b)
`B=420.855\ T`

(c)
`f=11.7647* 10^(14)\ Hz`

(d)
`T=8.5* 10^(-14)s`

Step-by-step explanation:

Given:

Kinetic Energy of an electron,
`KE=1.4\ keV=1400\ eV=1400* 1.6* 10^(-19)\ J`

radius of the orbit,
`r=0.3\ m`

we have:

mass of an electron,
`m=9.109* 10^(-31)\ kg`

charge on an electron,
`q=1.6* 10^(-19)\ C`

(a)

we know:

`KE=(1)/(2) m.v^2`

`1400* 1.6* 10^(-19)=0.5* (9.109* 10^(-31))* v^2`

`v=22.177* 10^(12)\ m.s^(-1)`

(b)

We also have the relation after the comparison of forces(centripetal and magnetic) on a moving charge in a magnetic field as:

`m.v =q.B.r` ...........................(1)

where:

B = magnetic field normal to the plane of circulating charge

putting respective values in eq. (1)

`(9.109* 10^(-31))* (22.177* 10^(12))=(1.6* 10^(-19))* 0.3* B`

`B=420.855\ T`

(d)

angular speed:

`\omega=(v)/(r)`

`\omega=(22.177* 10^(12))/(0.3)`

`\omega=73.92* 10^(12)\ rad.s^(-1)`

∴Time taken for 1 radian:

`t=(1)/(\omega)`

`t=(1)/(73.92* 10^(12))\ s.rad^(-1)`

Now time take for 1 circulation i.e. 2π radians(Time period):

`T=2\pi* t`

`T=2\pi* (1)/(73.92* 10^(12))`

`T=8.5* 10^(-14)s`

(c)

we know frequency :

`f=(1)/(T)`

`f=(1)/(8.5* 10^(-14))`

`f=11.7647* 10^(14)\ Hz`