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An electron moves at 2.80 ×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.60 ×10−2 T. Part A What is the largest possible magnitude of the acceleration of the electron due to the magnetic field? Express your answer with the appropriate units. aa = nothing

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Answer:

[tex]a = 3.73*10^{16}m/s^2[/tex]

Explanation:

The magnetic force is given by the equation,

[tex]\vec{F}_{magnetic} = q (\vec{v}X\vec{B})[/tex]

[tex]\vec{F}_{magnetic} = qvBsin\theta[/tex]

Where [tex]\theta[/tex] is the angle between velocity vector and the magnetic field,

That angle is 90°.

We know as well that

F=ma, replacing the mass and the acceleration in our previous equation we have

[tex]ma = qvBsin\theta[/tex]

[tex]a= \frac{qvBsin\theta}{m}[/tex]

Our values are,

[tex]q= 1.6*10^{-19}c[/tex]

[tex]v=2.8*10^6m/s[/tex]

[tex]B=7.6*10^{-2}T[/tex]

[tex]m=9.11*10^{-31}kg[/tex]

Substituting,

[tex]a = \frac{(1.6*10^{-19})(2.8*10^6)(7.6*10^{-2})}{9.11*10^{-31}}[/tex]

[tex]a = 3.73*10^{16}m/s^2[/tex]

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