An electron travelling at 3.3 x10^7 m/s perpendicular to a magnetic field of 0.1

Question

An electron travelling at 3.3x10^7 m/s perpendicular to a magnetic field of 0.16T moves in a circular path.

a) What is the magnitude of the magnetic force acting on the electron?

b) What is the radius of the electron's circular path?

c) Why does the electron move in a circular path?

( ) because the direction of the magnetic force does not change

( ) because the electron is not accelarating

( ) because the magnetic force acting on the electron is parallel to the electron's velocity

( ) because the magnetic force acting on the electron is perpendicular to the electron's velocity

( ) because the magnetic force acting on the electron is perpendicular to the electron's accelaration

Explanation / Answer

a) F = q v B = 1.6E-19*3.3E7*0.16=8.45E-13 N

b) q v B= mv^2/r

r = mv /q B = 9.11E-31*3.3E7/(1.6E-19*0.16)=1.17E-3 m

c) because the magnetic force acting on the electron is perpendicular to the electron's velocity

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