Question

Problem 18: An electrically charged base ball of mass 150g travels with a constant velocity of 89.5mph on a horizontal straight line in a southwestern direction, suspended against gravity by the magnetic force due to the earth's magnetic field. The earth's B-field is pointing straight north, horizontally, with a strength of 60 T. The earth's gravitational acceleration is 9.81m/s2 and 1mile = 1,609m. Neglect air resistance. Find y, the electric charge (amount and sign!). carried by the ball. Hints: (1) Draw the ball's velocity vector, T, and the magnetic field vector, B, in a a top- down view of the horizontal plane (= the plane containing the directions North, South, East and West, see: https://tinyurl.com/ydyjpqva). (2) What is the angle enclosed between vectors 7 and B? (3) What is the direction of the gravitational force vector (weight force), relative to the horizontal plane of your drawing? (4) What is the direction of the cross- product vector ūXB, relative to the horizontal plane of your drawing? (5) What would be direction of the magnetic force acting on the ball if you assume, for example, q> 0? (A) 4 = (B) q = (C) q = (D) q = (E) 4 = 867, C +867. C -434. C +134. C -613.

Answer 1

Given u = 89.5mpha 40.01 m/s stuswey fu ms150x10% kg n Bascut .. 1 .- ---E u-01 Almg Ahme, quen v= 29.5 mpha 40-mais by Lorentz Force Fu qlüxB) -O by newton's second have - mgii) so from e90 & 90 v=89.suph 's uu 9(TXB) - mg (-3) where = oloso (f) + Usino-f) ū = 40.01603 45° -9) + 40-01 Sinusº(-3) B = 60X166 ($) so Tex B= (40-01/03 45 + 40-01 Sin 45 €8]] x 60 x10 of fxf = 0 -1x$c- in cross froduct TxB = (40.01) (60 x169) Cesuse (*) 7 (40-01X6oxlobal ) * = |soxmox9-811-f ) 150*9.81 X10 40.01x60x1õ6 % = 8670 option 'B' is correct