An electron moves at 3.00×10 6 m/s through a region in which there is a magnetic

Question

An electron moves at 3.00×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 8.00×102 T .

Part A

What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?

Part B

What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field?

Part C

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

F = Q v B sin = m a , a = Q v B sin / m … where … 0 sin 1 … it follows that …
… a (min) = 0 … and … a (max) = Q v B / m =4.21*10^16

a = ¼ a (max) = ¼ ( Q v B / m ) = Q v B sin / m calculate theta by putting values

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