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 = -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 :
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)