An electron moves at 2.90×10 6 m/s through a region in which there is a magnetic

Question

An electron moves at 2.90×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.00×10?2 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

a)

Here magnetic force is balanced by force due to acceleration

qvBsin(o) =ma

=>a=qvBsin(o)/m

Here acceleration is maximum when o=90o

amax=(1.6*10-19)(2.9*106)(0.07)*sin90/(9.11*10-31)

amax=3.565*1016 m/s2

b)

acceleration is minimum when o=0o

=>amin=qvBsin(0)/m =0 m/s2

c)

given a=3.565*1016/4=8.913*1015 m/s2

=>8.913*1015=(1.6*10-19)(2.9*106)(0.07)sin(o)/(9.11*10-31)

o=14.48o

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