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nfluenced by magnetic An electron moves following a circular path in vacuum i field. The radius of the electron's path is 2.0 x 102 m, magnetic field strength is 2.0x 10-2 T, and specific charge electron e/m is-I .76 x 1011 C kg" (b) i. Write an expression for angular velocity of electron. ii. Calculate its orbital period ii. What happen to the electron's path when its velocity decreases? [10 marks]

Answer 1

Given data:-

Radius of circular path ,r= 2.0×10²m

Magnetic field,B=2.0×10^-2T

Specific charge of electron,$\frac{e}{m}$ =1.7×10¹¹C/kg

Let velocity of electron is v

We know force on charge particle moving in magnetic field ,

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

As path is circular so velocity is perpendicular to magnetic field .and magnetic field provides required value of centripetal force to move on circular path.

So, F= evB

Centripetal force = mv²/r

Thus evB= mv²/r

Solving we get , v/r =(e/m)×B

We know angular velocity = v/r

So I) angular velocity expression =( e/m)×B

ii) orbital period= 2πr/v = 2π/((e/m)×B)

Putting value of e/m= 1.76×10¹¹C/kg and B =2×10^{​​​​​-2}T

We get orbital period =1.78×10^-9 sec

iii) we know r= (mv)/(eB)

So with decrease in velocity radius of circular path decrease.