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 7.00×10⁻² 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 14 of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

Answer 1

a)

F = qVB = 1.6* 10^-19 * 2.6* 10^6* 7* 10^-2

9.11* 10^-31* a= 2.912* 10^-14 N

a = 3.196* 10^16 m/s^2

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b)

minimum possible acceleration

a = 0 m/s^2

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c)

a = 0.25* 3.196 = 8* 10^15 m/s^2

now

ma = F sin x

9.11* 10^-31* 8* 10^15 = 2.912* 10^-14 sin x

x = 14.5

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