Answer to Question #172063 in Physics for Ashley

Question #172063

A particle has a charge of q = +5.1 μC and is located at the origin. As the drawing shows, an electric field of Ex = +214 N/C exists along the +x axis. A magnetic field also exists, and its x and y components are Bx = +1.2 T and By = +1.5 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.



1
Expert's answer
2021-03-23T08:15:28-0400

a)


FBx=FBy=0 NFE=(5.1106)(214)=1.09103 NF_{Bx}=F_{By}=0\ N\\F_E=(5.1\cdot10^{-6})(214)=1.09\cdot10^{-3}\ N


Direction: along the + x axis

b)


FE=(5.1106)(214)=1.09103 NF_E=(5.1\cdot10^{-6})(214)=1.09\cdot10^{-3}\ N

Direction: along the + x axis


FBx=0F_{Bx}=0FBy=(5.1106)(1.5)(345)=2.64103 NF_{By}=(5.1\cdot10^{-6})(1.5)(345)=2.64\cdot10^{-3}\ N

Direction: along the + z axis

c)


FE=(5.1106)(214)=1.09103 NF_E=(5.1\cdot10^{-6})(214)=1.09\cdot10^{-3}\ N


Direction: along the + x axis


FBy=(5.1106)(1.5)(345)=2.64103 NF_{By}=(5.1\cdot10^{-6})(1.5)(345)=2.64\cdot10^{-3}\ N

Direction: along the - x axis

FBx=(5.1106)(1.2)(345)=2.11103 NF_{Bx}=(5.1\cdot10^{-6})(1.2)(345)=2.11\cdot10^{-3}\ N

Direction: along the + y axis



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