Oxygen Nucleus (Q=+8e) Moving in a B Field Part A An nucleus (charge ) moving ho

Question

Oxygen Nucleus (Q=+8e) Moving in a B Field Part A An nucleus (charge ) moving horizontally from west to east with a speed of 500 km/s experiences a magnetic force of 0.00320 nN vertically downward. Find the magnitude of the weakest magnetic field required to produce this force. Part B What is the direction of the weakest magnetic field required to produce this force. Part C An electron moves in a uniform, horizontal 2.10-T magnetic field that is toward the west. What must the magnitude of the minimum velocity of the electron be so that magnetic force on it will be 4.60 pN, vertically upward? Part D What is the direction of the minimum velocity?

Explanation / Answer

F = qV x B.. if V is in +x direction, for force to be in -Z direction, Force is to be in -Y direction... Since cross product of +x and -y is -Z... B = 0.00320 x 10^-9 N / 8e C x 500 x 10^3 m/sec = 5 Tesla.. Case 2: if B is in -x direction, for force to be in Z direction, Velocity is to be in -Y direction... charge of electron is -ve.. hence qV will be in +Y direction.. Since cross product of y and -x is Z... V = 4.6 x 10^-12 N / e x 2.1 = 27.38 m/sec... If numericals were wrong, plz rate me for the procedure.. Thank you..

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