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A particle with mass [tex]1.81 \times 10^{-3}\;{\rm kg}[/tex] and a charge of [tex]1.22 \times 10^{-8}\;{\rm C}[/tex] has, at a given instant, a velocity [tex]\vec{v} = (3.00 \times 10^{4}\;{\rm m}/{\rm s})\hat{j}[/tex].

What is the magnitude of the particle's acceleration produced by a uniform magnetic field [tex]\vec{B}=(1.63\;{\rm T})\hat{i} + (0.980\;{\rm T})\hat{j}[/tex]?

I use the equation F = qv (CROSS) B.

I rewrite this as F = q*v*B*sin().

I plug in to get F = (1.22*10^(-8))*(3.00*10^(4))*(1.90) and get (6.96*10^(-4)).

I divide this by the mass, but my answer is wrong.

Obviously my procedure is not correct. What's up?

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# Homework Help: Magnetic Fields and Acceleration

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