Question

Question 4.11: In a chamber, a uniform magnetic field of 6.5 G (1 G = 10 –4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10 6 m s –1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10 –19 C, me= 9.1×10 –31 kg)

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-169

Answer 1

$\text{the magnetic force,f=qvB act normal}\\\text{to the direction of motion,thus provide}\\\text{the necessary centripetal force to follow}\\\text{the circular path.}\\\\e.g.,f=qvB=\frac{mv^2}{r}\\\\r=\frac{mv}{qB}\\\\=\frac{9.1\times10^{-31}\times4.8\times10^6}{1.6\times10^{-19}\times6.5\times10^{-4}}\\\\=4.2\times10^{-2}=42mm$

Answer 2

Answer:

In a chamber, a uniform magnetic field of $6.4*10^{-4}T$ is maintained. An electron is shot into the field with a speed of $4.8*10^{6}m/s$ normal to the field. The path will be circular since the force on electron is perpendicular and the radius of the circular orbit is $4.2*10^{-2} m$

Explanation:

When a charged particle with charge $q$ enters in a uniform magnetic field of intensity $B$ with velocity $v$ a magnetic force called Lorentz force will act on the particle, that is given by

$F=q\left(v\times B\right)$

the direction of the force will be perpendicular to both velocity and the magnetic field. since force is the cross-product of velocity and field.

so it follows a circular path and the necessary centripetal force is provided by the Lorentz force.

to get the radius of the circular path we need to equate both forces

$F=q\left(v\times B\right)=\frac{mv^2}{r}$

from the question, $q=1.6*10^{-19}C, v=4.8*10^{6} m/s, m=9.1*10^{-31}kg,B=6.5*10^{-4} T$

substitute these values to get radius,$r=\frac{mv}{qB} \\r=\frac{9.1*10^{-31}*4.8*10^{6} }{1.6*10^{-19}*6.5*10^{-4} }=4.2*10^{-2} m$