A particle has a charge of q 53 mu C and is located at the

A particle has a charge of q = +5.3 mu C and is located at the origin. As the drawing shows, an electric field of E_x = +211 N/C exists along the +x axis. A magnetic field also exists, and its x and y components are B_x = + 1.6 T and B_y = + 1.6 T. Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is stationary, moving along the +x axis at a speed of 345 m/s, and moving along the +z axis at a speed of 345 m/s.

Solution

(a) if the particle is stationary then the force due to magnetic field is zero and only electric force work on it

Fe=qE=5.3*10^-6 C*211N/C=1118.3*10^-6 N=1.18*10^-4N (positive x direction)

(b)particle is moving along the positive x axis with velocity 345m/sec

force due to Bx=0(because Fm= qvbsin(theta) here theta=0 because particle is moving along x axis

force due to By=5.3*10^-6 C*345m/sec*1.6T=2925.6*10^-6=29.25*10^-4 N (along z axis)

force due to electric field is same as in part (a) because electric force dont affect due to motion of charge

so the total force=sqrt( (1.18)² +(29.25)² )*10^-2 =29.27*10^-2 N along y axis

(c) particle is moving along z axis

so force due to Bx=qvB= 29.25*10^-4 N(y direction)

force due to By=qvB= 29.25*10^-4 N (x direction)

resultant Fb=sqrt{(29.25)² +(29.25)² }*10^-2 =41.36*10^-2 Nalong z axis

Fe=1.18*10^-4 N along x axis

so total F=sqrt(1.18)² +(41.36)² *10^-2 =41.37*10^-2 N along y axis