Question

What is the value of e/m for a particle that moves in a circle of radius 13 mm in a 0.82-T magnetic field if a perpendicular 630-V/m electric field will make the path straight? Express your answer using two significant figures

Answer 1

here,

the magnetic field strength , B = 0.82 T

the electric field strength , E = 630 V/m

let the velocity of particle be v

when it moves in straight line

q * E = q * v * B

v = E /B = 630 /0.82 m/s

v = 768.3 m/s

the radius of the circle , r = 13 mm = 0.013 m

0.013 = m * v /(q * B)

0.013 = (m * 768.3)/(q * 0.82)

solving for (q/m)

(q/m) = 7.2 * 10^4 C/kg