A proton having a velocity of magnitude 3.45 times 10^4 m/s passes into a region

Question

A proton having a velocity of magnitude 3.45 times 10^4 m/s passes into a region of uniform magnetic field at an angle of alpha = 30.0 degree with respect to the horizontal x-axis as shown in the figure below. The magnetic field has a magnitude of B = 0.212 T and is pointing out of the paper in the negative y-direction so that its vector is given by the expression B vector = 0.212T(-y^). (a) Use the right hand rule to find the direction of force on the proton as it passes into the region where the magnetic field is uniform. Sketch the force direction relative to the initial direction of v vector. Explain how you applied the rule. (b) Show that the vector representing the proton's velocity is given by: v vector = 2.99 times 10^4 m/s(+x^)+ 1.73 times 10^4 m/s(+z^) (c) Use the vector cross product (mathematical form) to find the vector representing the force on the proton just as it enters the region of uniform magnetic field. You may want to review the relevant section in t

Explanation / Answer

a)The force is perpendicular to both the velocity and the magnetioc field vectors. So the force will directed along -z.

b) velocity along x will be:

Vx = V sin(alpha)

Vx = 3.45 x 10^4 x sin(60) = 2.99 x 10^4 m/s

Vy = 3.45 x 10^4 x cos(60) = 1.73 x 10^4 m/s

V = (2.99 x 10^4 m/s) i + (1.73 x 10^4 m/s) j

c)We know that the magnetic force on a moving charge is:

F = q (v x B)

F = 1.6 x 10^-19 [((2.99 x 10^4 m/s) i + (1.73 x 10^4 m/s) j) x (-0.212 j) ]

F = 1.6 x 10^-19 (0 i + 0j - 6.34 x 10^3 k)

F = -1.01 x 10^-15 N k

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