Question

Tin- diagram below shows a particle of charge q_0 = 2.00 muC moving along the +y axis with a speed v = 4.00 times 10^6 m/s. A magnetic field B of magnitude 4 00 times 10^-5 T is directed the +Z axis, and an electric field E of magnitude 120 N/C points along the -k axis Determine The magnitude and direction of the magnetic force on the charge. The magnitude direction of the electric force on the charge. The magnitude and direction of the net force that acts the charge. The value to which the magnitude of the electric field would have to be adjusted so that here would be no net force on the charge.

Answer 1

(a) The magnitude and direction of magnetic force on the charge will be given as :

using a formula, we have

F_B = q₀ v B sin $\theta$ = (2 x 10^-6 C) (4 x 10⁶ m/s) (4 x 10^-5 T) sin 90⁰

F_B = 3.2 x 10^-4 N

(b) The magnitude and direction of electric force on the charge will be given as :

we know that, F_e = q₀ E

F_e = (2 x 10^-6 C) (120 N/C)

F_e = 2.4 x 10^-4 N

(c) The magnitude and direction of net force that acts on the charge will be given as :

F_net = F_B + F_e = (3.2 x 10^-4 N) + (2.4 x 10^-4 N)

F_net = 5.6 x 10^-4 N

(d) The value to which the magnitude of an electric field will be given as :

F_e = F_B

q₀ E = q₀ v B

E = v B = (4 x 10⁶ m/s) (4 x 10^-5 T)

E = 160 N/C