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

Answer:

The potential difference and the magnetic field are 200.655 V and $9.55\times10^{-4}\ T$

Explanation:

Given that,

Speed $v=8.4\times10^{6}\ m/s$

radius = 5.0 cm

We need to calculate the potential difference

We know that,

The kinetic energy acquired by electron

$\dfrac{1}{2}mv^2=q\Delta V$

$\Delta V=\dfrac{1}{2}\times\dfrac{mv^2}{q}$

Where, m = mass of  electron

q = charge

v = velocity

Put the value into the formula

$\Delta V=\dfrac{9.1\times10^{-31}\times(8.4\times10^{6})^2}{2\times1.6\times10^{-19}}$

$\Delta V=200.655\ V$

(b). We need to calculate the magnetic field

Using formula of magnetic field

$B=\dfrac{mv}{qr}$

Put the value into the formula

$B=\dfrac{9.1\times10^{-31}\times8.4\times10^{6}}{1.6\times10^{-19}\times5.0\times10^{-2}}$

$B=9.55\times10^{-4}\ T$

Hence, The potential difference and the magnetic field are 200.655 V and $9.55\times10^{-4}\ T$

Answer 2

Answer:

a) 200 v

b )9.555 x 10⁻⁴ T

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 = $\frac{mv}{qB}$

B )  magnetic field required =

$\frac{mv}{qr}$

= 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