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devlian [24]
3 years ago
9

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
Physics
1 answer:
d1i1m1o1n [39]3 years ago
4 0

Answer:

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

Explanation:

The magnetic force is given by the equation,

\vec{F}_{magnetic} = q (\vec{v}X\vec{B})

\vec{F}_{magnetic} = qvBsin\theta

Where \theta 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

ma = qvBsin\theta

a= \frac{qvBsin\theta}{m}

Our values are,

q= 1.6*10^{-19}c

v=2.8*10^6m/s

B=7.6*10^{-2}T

m=9.11*10^{-31}kg

Substituting,

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

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

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