Question

The electrons in the beam of an old-fashioned television tube have an energy of 12 keV. The tube is oriented so that the electrons move horizontally from south to north. The vertical component of the earth's magnetic field points down and has magnitude . (a) In what direction will the beam deflect? (b) What is the acceleration of a given electron? (c) How far will the beam deflect in moving 20 cm through the TV tube?

Answer 1

Part a)

force will be along East direction

Part b)

`a = 6.27 * 10^(14) m/s^2`

Part C)

`x = 2.96 * 10^(-3) m`

Step-by-step explanation:

As we know that vertical component of the magnetic field is given as

`B_y = 55 \mu T`

Kinetic energy of the electron beam is given as

`K = 12 keV`

Part a)

As we know that force on moving charge due to magnetic field is given as

`\vec F = q(\vec v * \vec B)`

now we know that

`\vec F = (-e)(v \hat j * B (-\hat k))`

`\vec F = evB \hat i`

so force will be along East direction

Part b)

As we know that

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

`(1)/(2)mv^2 = 12 * 10^3 (1.6 * 10^(-19)) J`

`v = 6.5 * 10^7 m/s`

now acceleration of the charge is given as

`a = (evB)/(m)`

`a = ((1.6 * 10^(-19))(6.5 * 10^7)(55 * 10^(-6)))/(9.11 * 10^(-31))`

`a = 6.27 * 10^(14) m/s^2`

Part C)

times taken by the beam of electron to move through the distance of 20 cm is given as

`t = (L)/(v)`

`t = (0.20)/(6.5 * 10^7)`

`t = 3.08 * 10^(-9) s`

so the deflection in the direction of magnetic field is given as

`x = (1)/(2)at^2`

`x = (1)/(2)(6.27 * 10^(14))(3.08 * 10^(-9))^2`

`x = 2.96 * 10^(-3) m`