A magnetic field has a magnitude of 1.2x10^-3 T, and an electric field has a mag

Question

A magnetic field has a magnitude of 1.2x10^-3 T, and an electric field has a magnitude of 4.6x10^3 N/C. Both fields point in the same direction. A positive 1.8x10^-6 – C charge moves at a speed of 3.1x10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge

i tried the equation F=charge(q)*velocity*magnetic field(B)*sin90. and that didnt work. That equation didnt include the electric field though so i think i need to incorporate that somehow. So if you could explain please that would be great :)

Explanation / Answer

Assume that Electric field and magnetic field point along positive x axis

E = 4.6x103 N/C

Electric force on positive charge have same direction with the electric force so

FE = q.E = 1.8x10-6.(4.6x103)= 8.28x10-3 N along positive x axis

Assume that velocity of the charge along positive y axis

Magnetic force on the charge with right hand rule is point along negative z axis

FB = q.v.B sin 900 = 1.8x10-6 x 3.1x106 x 1.2x10-3 = 6.696x10-3 N

Net force on charge :

FR = (FE2 + FB2) = 10.649 x 10-3 N

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