help An electron moves at 2.90 times 106m/s through a region in which there is a

Question

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An electron moves at 2.90 times 106m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.70 times 10-2T. What is the largest possible magnitude of the acceleration of the electron due to the magnetic field? What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field? If the actual acceleration of the electron is 1/4 of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

Explanation / Answer

A.)When magnetic field is perpendicular to the velocity, acceleration is maximum

So ma=qvBsin(90)

a=qvB/m=1.602 x 10^-19*2.9x10^6*7.7 x 10^-2/9.1 x 10^-31 =3.93 x 10^16 m/s^2

B.)When magnetic field is parallel to the velocity, acceleration is minimal

So ma=qvBsin(0)

ma=0

So a=0 m/s^2

C.)3.93/4 x 10^16=1.602 x 10^-19*2.9x10^6*7.7 x 10^-2*sin(theta)/9.1 x 10^-31

So theta=sin^-1(0.25)=14.48 degrees

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