Question

Consider an electron being ejected from the surface of a heated filament at nearly zero specd [a] Through what potential difference must it be accelerated to achieve a speed of 8.4x106m/s? Any relativistic effects can be neglected [b] What is the minimum B-field required to cause the electron in part [a] above to move in a circular orbit of radius r 5.0 cm?

Answer 1

a) 200 v

b )9.555 x 10⁻⁴ T

Step-by-step explanation:

Equation for electron acceleration under potential difference

q V = 1/2 mv²

V is potential diff , v is velocity after acceleration, q is charge on electron.

V= .5 x 9.1x 10⁻³¹ x( 8.4 x 10⁶)² / 1.6 x 10⁻₁¹⁹

= 200.655 V

b ) radius of circular orbit in a magnetic field of electron

r =
`(mv)/(qB)`

B ) magnetic field required =

`(mv)/(qr)`

= 9.1 x 10⁻³¹ x 8.4 x 10⁶ / 1.6 x 10⁻¹⁹ x 5 x 10⁻²

9.55 x 10⁻⁴ T

Answer 2

The potential difference and the magnetic field are 200.655 V and
`9.55*10^(-4)\ T`

Step-by-step explanation:

Given that,

Speed
`v=8.4*10^(6)\ m/s`

radius = 5.0 cm

We need to calculate the potential difference

We know that,

The kinetic energy acquired by electron

`(1)/(2)mv^2=q\Delta V`

`\Delta V=(1)/(2)*(mv^2)/(q)`

Where, m = mass of electron

q = charge

v = velocity

Put the value into the formula

`\Delta V=(9.1*10^(-31)*(8.4*10^(6))^2)/(2*1.6*10^(-19))`

`\Delta V=200.655\ V`

(b). We need to calculate the magnetic field

Using formula of magnetic field

`B=(mv)/(qr)`

Put the value into the formula

`B=(9.1*10^(-31)*8.4*10^(6))/(1.6*10^(-19)*5.0*10^(-2))`

`B=9.55*10^(-4)\ T`

Hence, The potential difference and the magnetic field are 200.655 V and
`9.55*10^(-4)\ T`