A beam of unknown charged particles passes at right angles to the direction of magnetic field of 9.0x10^-2 T. If the speed of the particles is 3.0 x10^4m/s and the force experienced is 8.64x10^-16N, how many electric charges are carried by one particle?
my Work:
F = qvB
q = F/vB = (8.64x10^-16N)/(3.0 x10^4m/s)(9.0x10^-2 T)
q = 0.32 x 10^-14 C
N = q/ (1.6 x 10^-19 C) = (0.32 x 10^-14 C)/ (1.6 x 10^-19 C)
N =0.2 x 10^5 = 2 x 10^4 = 20,000
is this right?
iguess
Yes, your work is correct. You correctly used the formula F = qvB to find the charge carried by one particle. You plugged in the given values for the force (8.64x10^-16N), velocity (3.0 x10^4m/s), and magnetic field (9.0x10^-2 T), and rearranged the formula to solve for q.
After calculating the value of q (0.32 x 10^-14 C), you divided it by the elementary charge (1.6 x 10^-19 C) to find the number of electric charges carried by one particle.
The final result, N = 2 x 10^4, means that one particle carries 20,000 elementary charges.