The figure shows a particle that carries a charge of q_0 = -2.59 Times 10^-6 C.

Question

The figure shows a particle that carries a charge of q_0 = -2.59 Times 10^-6 C. It is moving along the +y axis at a speed of v = 1.91 Times 10^6 m/s. A magnetic field B rightarrow of magnitude 3.29 Times 10^-5 T is directed along the +z axis, and an electric field E rightarrow of magnitude 116 N/C points along the -x axis. Determine (a) the magnitude and (b) direction (as an angle within x-y plane with respect to +x-axis in the range (-180 degree, 180 degree]) of the net force that acts on the particle. Number units Number units

Explanation / Answer

electric force=q*E

as charge is negative, force will be in opposite direction of electric field

hence electric force is along +ve x axis.

electric force magnitude=2.59*10^(-6)*116=3.0044*10^(-4) N

magnetic force=q*(cross product of v and B)

here velocity is along +ve y axis and B is along +ve z axis.

their cross product will be +ve x axis

as charge is negative, magnetic force will be along -ve x axis.

magnetic force magnitude=2.59*10^(-6)*4.91*10^6*3.29*10^(-5)

=4.1839*10^(-4) N

so total force is along -ve x axis.

total force magnitude=4.1839*10^(-4)-3.0044*10^(-4)=1.1795*10^(-4) N

angle with +ve x axis=180 degrees

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