help A proton moving at 4.30 times 106 m/s through a magnetic field of magnitude

Question

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A proton moving at 4.30 times 106 m/s through a magnetic field of magnitude 1.64 T experiences a magnetic force of magnitude 8.80 times 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest.) Assume the region to the right of a certain plane contains a uniform magnetic field of magnitude b = 1.39 mT and the field is zero in the region to the left of the plane as shown in the figure below. An electron, originally traveling perpendicular to the boundary plane, passes into the region of the field. Determine the time interval required for the electron to leave the "field-filled" region, noting that the electron's path is a semicircle. S Assuming the maximum depth of penetration into the field is 1.99 cm, find the kinetic energy of the electron. eV

Explanation / Answer

sin(theta) = F/(e v B) = 8.8e-13/(1.6e-19*4.30e6*1.64) = 0.77992

===> theta = 51.3 degrees

===> theta = 231 degrees

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a)

q v B = m v^2/R

==> R = m v/qB

t = pi R/v = pi m/qB = 3.1416*9.109e-31/1.6e-19/1.39e-3 = 1.29x 10^-8 s

b)

R = 1.99e-2 m

R = m v/qB

===> v = qBR/m = 1.6e-19*1.39e-3*1.99e-2/9.109e-31 = 4.859e6 m/s

===> K = 0.5 m v^2 = 67.2 eV

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