Question

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The plates of a Thomson q/m apparatus are 6.0 cm long and are separated by 1.0 cm. The end of the plates is 30.0 cm from the tube screen. The kinetic energy of the electrons is 2.2 keV.

(a) If a potential of 25.0 V is applied across the deflection plates, by how much will the beam deflect?

_______________ mm

(b) Find the magnitude of the crossed magnetic field that will allow the beam to pass through undeflected.

_______________ T

Answer 1

Part a).

As Potential,V=25 V

sepration between plates,d=1 cm=0.01 m

so magnitude electric field,E=V/d=25/0.01=2500 V/m

kinetic energy of the electrons is,K.E= 2.2 keV

so velocity of electron.v=((2 $\times$ K.E)/m)^1/2=((2 $\times$ 2.2 $\times$ 10³ $\times$ 1.6 $\times$ 10^-19)/(9.1 $\times$ 10^-31))^1/2=27814139.81 m/s

horizontal distance is given by

x=vt

so t=x/v

and deflection y is given by

y=(1/2)at²=(1/2)(eE/m)(x/v)²=(1/2)((1.6 $\times$ 10^-19 $\times$ 2500)/(9.1 $\times$ 10^-31))(0.36/(27814139.81))²=0.03681 m=36.81 mm

deflection is=36.81 mm

Part b)

when beam passes straight

eE=evB

B=E/v=2500/(27814139.81)=8.99 $\times$ 10^-5 T

magnitude of the crossed magnetic field=8.99 $\times$ 10^-5 T