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Chapter 28: Q 8P (page 829)

Question: An electric field of1.50kV/mand a perpendicular magnetic field of 0.400Tact on a moving electron to produce no net force. What is the electron’s speed?

Short Answer

Expert verified

The electron’s speed is |v|=3.75x103m/s.

Step by step solution

01

Given

ϕ=90.0∘E=1.50kVm=1.50x103V/mB=0.400TF=0

02

Determining the concept

Find the velocity of an electron by inserting the given values in the formula for electromagnetic force.

Formulae are as follow:

F=eE+evBsinϕ

Where, F is force, v is velocity, E is electric field, B is magnetic field, e is charge of particle.

03

Determining the speed of electron

The electromagnetic force experienced by an electron is,

F=eE+evBsinϕ

Inserting given values,

0=-1.6×10-19×1.50×103+-1.6×10-19×v×0.400×sin90∘0=-1.6×10-19×1.50×103-1.6×10-19×v×0.400×sin90∘v=1.50×1030.400v=3.75×103m/s

Hence, the electron’s speed is v=3.75×103m/s.

Therefore, the velocity of an electron can be determined by inserting the given values in the formula for electromagnetic force.

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