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

its c i just checked it