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inysia [295]
3 years ago
15

An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (

5.5T)i – (3.7T)j. What is the acceleration of the electron when it first enters the region of the uniform magnetic field?
Physics
1 answer:
Crazy boy [7]3 years ago
5 0

Answer:

Acceleration, a=9.36\times 10^{18}\ m/s^2

Explanation:

It is given that,

Speed of electron, v=8\times 10^6\ m/s

Charge on an electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

Magnetic field, B=5.5i-3.7j

Magnitude, |B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T

Magnetic force is given by :

F=qvB

Also, F = ma

a=\dfrac{qvB}{m}

a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}

a=9.36\times 10^{18}\ m/s^2

So, the acceleration of the electron is 9.36\times 10^{18}\ m/s^2. Hence, this is the required solution.

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