A particle has a charge of q = +5.6 C and is located at the origin. As the drawi

Question

A particle has a charge of q = +5.6 C and is located at the origin. As the drawing shows, an electric field of Ex = +212 N/C exists along the +x axis. A magnetic field also exists, and its x and y components are Bx = +1.5 T and By = +1.7 T. Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the +x axis at a speed of 345 m/s, and (c) moving along the +z axis at a speed of 345 m/s.

Explanation / Answer

Solution,

A)

Force, F = q v B sin(theta)

Since the particle is at rest it will experience only electric force not magnetic force.

F_E = eE = (5.6 x 10^-6) (212) = 1.187 x 10^-3 N

(along +ve x direction)

F_Bx = 0

F_By = 0

B)

Given velocity, v = 345 m/s

F_E = eE = (5.6 x 10^-6)(212) = 1.187 x 10^-3 N (along +ve x direction)

F_Bx = qvB sin0= 0 N

F_By = qvB sin90= (5.6 x 10^-6)(345)(1.7) = 3.28 x 10^-3 N

C)

F_E = (5.6 x 10^-6) x 212 = 1.187 x 10^-3 N (along +ve x direction)

F_Bx = qvBx sin90 = 2.898 x 10^-3 N positive y axis

F_By = qvBsin90^0 =(5.6 x 10^-6)(345) (1.7) = 3.28 x 10^-3 N negative x axis

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