# 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.

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Expert Answers

thilina-g | Certified Educator

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.**