An electron is accelerated by a potential difference of 3000v and enters a region of a uniform magnetic field. as a result the electron bends along a path with a radius of curvature of 26.0 cm. find the magnitude of the magnetic field. show work.

The election has energy of 3000 eV entering the magnetic field. 

converting to J, energy = qeV =1.6E-19*3000 where q=1.6E-19

Kinetic energy = 1/2mv^2 with m is mass of electron

1/2mv^2 = qeV

v = sqrt[1.6E-19*3000*2/9.1E-31] = 3.25E6

Magnetic force = eVB where B is magnetic field strength

Force = ma where a=radial acceleration=v^2/r 

evB = mv^2/r

B =  2*3000/0.26/3.25E6 = 7.1E-3T = 71Gauss

the answer is a. hope this .