Question

An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.00mT . If the speed of the electron is 1.50×10⁷ m/s , determine(a) the radius of the circular path.

Answer 1

Answer: The Radius of the circular path is 3.9 cm.

Explanation:

The magnitude of the magnetic field , B = 2.00mT = 2 ×10^-3 T

the speed of the electron , v , = 1.50 ×10^7 m/s

we also know that the mass of an electron m =9.1 ×10^-31 kg

and the charge q on an electron = 1.6 ×10^-19 C

Now, because the electron is moving in a circular path there is a centripetal force acting on it to prevent it from slipping off the magnetic field.

this Force, F, is given by :

F =$\frac{mv^²}{r}$  -equation1

We also know from Lorentz's Law , the force acting on a moving charge in a magnetic field is given by

F = qvBsinθ --equation 2

on combining equations 1 and 2 we get the expression for Radius,  r , of the circular path as :

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

Putting the respective values of m,v,B,q and θ , we get

r = 3.9 ×10^-2 m = 3.9cm

(To know more about Force on moving charges: brainly.com/question/17081554)