Question

An unknown particle moves in a straight line through crossed electric and magnetic fields with E = 1.5 kV/m and B = 0.034 T. If the electric field is turned off, the particle moves in a circular path of radius r = 2.7 cm.

Answer 1

E=1.5x10^3 N/C
B=0.034T
r=2.7x10^-2 m

Particle is moving in straight line with constant speed.
So electric and magnetic fields are perpendicular to each other as well as velocity of particle is perpendicular to both.
In this case, electric force(qE) on the electron should balance the magnetic force(qvB) on it.
qvB=qE
v=E/B

Now centripetal force(mv^2/r) required to move along circular path is provided by magnetic force (qvB)
qvB=mv^2/r
qB =mv/r
Putting v=E/B,
qB = mE/rB
q/m = E/rB^2 = 1.5x10^3 / (2.7x 10^-2 x 0.034^2) =48057439.06 Ckg^-1

Answer 2

assuming to find spped:

as E = Bc

spped V = E/B

v = 1500/0.034

v = 4.411*10^ 4 m/s

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Answer 3

apply mv^2/r = qvB

as E = Bc

spped V = E/B

v = 1500/0.034

v = 4.411*10^ 4 m/s

so charge to mass ratio is q/m = v/Br

q/m = 4.411e4/(0.034 * 0.027)

q/m = 4.805 *10^7 C/kg

so this particle must be either 1H1 or   4 He 2