Chapter 20, Problem 14P

To determine

velocity of beam of electrons when goes undeflected from electric and magnetic field and radius of electron when electric field is turned off.

Answer to Problem 14P

Solution:

Velocity = $1.02\times {10}^6$ m/s and radius = $0.77\times {10}^{-3}$ m

Explanation of Solution

Given:

Electric field $E=7.7\times {10}^3$ V/m; Magnetic field $B=7.5\times {10}^{-3}$ T

Formula used:

$v=\frac{E}{B}$ and $r=\frac{mv}{qB}$

Calculation:

Since beam of electrons is not deflected in Electric field and Magnetic field so

Electric Force = Magnetic Force

$qE=qvB$

$v=\frac{E}{B}$

$v=\frac{7.7\times {10}^3}{7.5\times {10}^{-3}}$

$1.02\times {10}^6$ m/s

When Electric field is switched off then electrons are moving perpendicular to magnetic field, so they perform circular motion,

so radius of curvature will be

$r=\frac{mv}{qB}$

$r=\frac{9.1\times {10}^{-31}\times 1.02\times {10}^6}{1.6\times {10}^{-19}\times 7.5\times {10}^{-3}}$

$r=0.77\times {10}^{-3}$ m

Calculation:

When magnetic field and electric field are crossed to each other and electron beam is undeflected then this arrangement can be used to select electrons of velocity out of a beam of electrons moving with different speed. This concept is called velocity filter or velocity selector.