Determine the velocity of a beam of electrons that goes undeflected when moving perpendicular to an electric and magnetic field. The electric field and the magnetic field are also perpendicular to each other and have magnitudes 7.7 x 10^3 V/m and 7.5 x 10^-3 T , respectively.
a. What is the radius of the electron orbit if the electric field is turned off?

r = 7.8 × 10 ⁻⁴ m

Explanation:

Given: Electric field E = 7.7 x 10^3 V/m, Magnetic field = 7.5 x 10^-3 T

as |E| = |V| x |B|

⇒  |V| = 7.7 x 10^3 V/m / 7.5 x 10^-3 T = 1.03×10⁶ m/s

Formula for this question is F = qE + qVB  

⇒ F = q VB, ( as Electric component is zero)

the acceleration a = v²/r  and also a = F/m   ( r=radius of curvature) so

so qVB/m = v²/r      

⇒ r = mV/qB

r= (9.11 × 10 ⁻³¹ kg x 1.03×10⁶ m/s) / (1.6 × 10 ⁻¹⁹ C x 7.5 x 10⁻³ T)

r = 7.8 × 10 ⁻⁴ m

the answer would most likely be c, because if it's a comical piece of information, it may likely not be true, if it's a site that allows anybody to edit the content, it could mean a five year old is typing the info, and if it's a tv show produced by a celeb, it may have a biased opinion.

gravitational energy

electrical energy

kinetic energy

ligh energy