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A particle with mass 1.81x10^3 kg and a charge of...

A particle with mass 1.81x10^3 kg and a charge of 1.22 x 10^8 C has, at a given instant, a velocity v=(3.00x10^4 m/s)j. What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field B=(1.63T)i + (0.980T)j?

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Answer #1
F=ma=q(U*B) a=q/m (U*B) =(1.22*10^-8)/1.81*10^-2 [(3*10^4)j *(1.63i+0.983j) a=(0.33)(-k) a=0.33 m/s²
answered by: Syed Eassa
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