Question

An electron entering Thomson’s e/m apparatus has an initial velocity (in horizontal direction only) of 4.0 x 10 6 m/s. In the lab there is a permanent horseshoe magnet of strength 12 mT, which you would like to use. (a) What electric field will you need in order to produce zero deflection of the electrons as they travel through the apparatus? (b) The length of nonzero E and B fields is 2.0 cm. When the magnetic field is turned off, but the same electric field remains, how far in the vertical direction will the electron beam be deflected over this length?

Answer 1

Answer:

a)E=48 x 10 ³ V/m

b)y=0.102 m

Explanation:

Given that

V= 4 x 10 ⁶ m/s

B= 12 mT

a)

We know that

E= V .B

E= 4 x 10 ⁶ x 0.012

E=48 x 10 ³ V/m

b)

We know that

F = E q  

F = m a

E q = m a

In horizontal direction

L= V t

t= L/V

t= 0.02 /(4 x 10 ⁶)  s

$t=\dfrac{0.02}{4\times 10^6}\ s$

$t=5\times 10^{-9}\ s$

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The charge on electron

$q=1.6\times 10^{-19}\ C$

E q = m a

a=Eq/m

$a=\dfrac{48\times 10^3\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}\ m/s^2$

$a=8.4\times 10^{15}\ m/s^2$

$y=ut+\dfrac{1}{2}at^2$

Given that u = 0

$y=\dfrac{1}{2}at^2$

$y=\dfrac{1}{2}\times 8.4\times 10^{15}(5\times 10^{-9})^2\ m$

y=0.102 m