Find the magnitude of the acceleration of the electron: An electron moves at a speed of 1.3 x 10^7 m/s at right angles to a magnetic field with...  a magnitude of 0.40 T.

Expert Answers

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This problem can be solved by the following expression.

F = qvB

Where, F - Force,

q - Charge

B - Magnetic field.

v -velcoity of electron

q = 1.6 x 10^(-19) C (charge of an electron)

B = 0.4 T

v = 1.3 x 10^7 ms-1

 

F = 1.6 x 10^(-19) x 0.4 x 1.3 x 10^7 N

F = 8.32 x 10^(-13) N

 

But we know F =ma

m - mass of electron = 9.109 x 10^(-31) kg

8.32 x 10^(-13) N = 9.109 x 10^(-31) kg x a

a = 9.133 x 10^17 ms-2

 

Therefore th magnitude of the acceleration is 9.133 x 10^17 ms-2.

 

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