Question

Your answer is partially correct. A magnetic field has a magnitude of 0.0018 T, and an electric field has a magnitude of 4.6 x 103 N/C. Both fields point in the same direction. A positive 1.5-C charge moves at a speed of 3.3 x 106m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge. BA E 9 90° Number ! Units N

Answer 1

Net force acting on the particle will be:

Fnet = Magnetic force(Fm) + Electric force(Fe)

now, magnetic force on a moving charge in magnetic field is given by:

Fm = q*VxB

here, q = charge = 1.5 uC = 1.5*10^-6 C

V = Velocity of particle = 3.3*10^6 m/s (+i)

B = Magnetic field = 0.0018 T (+j)

Fm = q*(V i)x(B j) = q*V*B k

Fm = (1.5*10^-6)*(3.3*10^6)*0.0018 k

Fm = 0.00891 k

Electric force on charge particle qill be:

Fe = q*E

given, E = electric field = 4.6*10^3 N/C (+j)

then, Fe = (1.5*10^-6)*(4.6*10^3) j

Fe = 0.0069 j

So, net force(Fnet) = 0.0069 j + 0.00891 k

magnitude of net force will be:

|Fnet| = sqrt(0.0069^2 + 0.00891^2)

|Fnet| = 0.011269 N

|Fnet| = 1.13*10^-2 N