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An electron travels at a speed of 9,811.933 m/s through a uniform magnetic field whose magnitude is 0.075 T. What is the magnitude of the magnetic force on the electron if its velocity vector and the magnetic field vector are make an angle of 18.407o ? Express your answer as the coefficient in front of 10^-19.

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The force on a moving particle can be written as:

[tex]\vec{F}=q(vX\vec{B}+\vec{E})[/tex]

Where X is the vectorial product. In our case, as there is only a magnetic field, we'll have:

[tex]F=qvBsin(\theta)[/tex]

Where theta is the angle between both vectors. Replacing our values we get:

[tex]F=-1.6*10^{-19}*9811.933*0.075*sin(18.407)=-3.7179*10^{-17}[/tex]

Considering we want only the magnitude (not the signal) and in terms of 10^(-19), our final answer is:

F = 371.79 * 10^(-19) N

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