A 30.0-g metal ball having net charge Q = 5.45 µC is thrown out of a window hori

Question

A 30.0-g metal ball having net charge Q = 5.45 µC is thrown out of a window horizontally north at a speed v = 19.8 m/s. The window is at a height h = 21.4 m above the ground. A uniform, horizontal magnetic field of magnitude B = 0.0100 T is perpendicular to the plane of the ball's trajectory and directed toward the west. (a) Assuming the ball follows the same trajectory as it would in the absence of the magnetic field, find the magnetic force acting on the ball just before it hits the ground. (Let the +x-direction be toward the north, the +y-direction be up and the +z-direction be east.) F = N (b) Based on the result of part (a), is it justified for three-significant-digit precision to assume the trajectory is unaffected by the magnetic field? Yes No Explain.

Explanation / Answer

magnetic force Fb = q*(v x B)

along horizontal

acceleration ax = 0

the velocity remains same

velocity before hitting the ground

vx = 19.8 i m/s

magnetic force Fby = q*(v x B) = 5.45*(19.8 i x -0.01 k)

Fby = 5.45*19.8*0.01 j = 1.07 j N

along vertical

initial velocity voy = 0

acceleration ay = -9.8 j m/s^2

displacement y = -21.5 m

y= voy*t + (1/2)*ay*t^2

-21.5 = 0 - (1/2)*9.8*t^2

t = 2.09 s

velocity before hitting the ground

vy = voy + ay*t

vy = 0 - 9.8*2.09 j

vy = -20.5 j m/s

magnetic force Fbx = q*(v x B) = 5.45*(-20.5 j x -0.01 k)

Fbx = 5.45*20.5*0.01 i = 1.11 i N

Fb = sqrt(Fbx + Fby^2) = 1.54 N

(b)

NO

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