An electron moves at 2.60×10 6 m/s through a region in which there is a magnetic

Question

An electron moves at 2.60×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.80×102 T .

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

a= ??? m/s^2

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

a= ??? m/s^2

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

magnetic force on moving charge is givn by F = qvB sin theta

where q is charge

v is its velocity

B is manetic fied

theta is the angle V and B

so for max accleration , theta = 90 deg

then F = ma = qVB

a = (1.6 e-19 * 2.6 e 6 * 7.8 e -2)/(9.11 e-31)

a = 3.56 e 16 m/s^2

-------------------------

for min accleration , theta = 0 deg

so a = 0

------------------------------------------------------

sin theta = 1/4

theta = sin^-1(0.25)

theta = 14.47 deg

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