Motion and Electromotive Force Due in 4 hours, 1 minute Two parallel metal ods f

Question

Motion and Electromotive Force Due in 4 hours, 1 minute Two parallel metal ods form a plane inclined at 20 e to the horizontal. The upper end of the r s are connected together by a esistoro s20 o melo er end of he rods are held in an elect call·insula in block They are separated horizontally by a distance of 20.0 cm. There is a uniform magnetic field of 0.100 T directed vertically upward. A metal bar of mass 0.10 g is sliding down the metal rods. What is the maximum speed of the bar? Assume: no frictional loss and that the rods are very long: g-9.81 m/s? Submit Answer Tries 0/10

Explanation / Answer

Using newton's law

Fnet = 0

FB + F = 0

where FB is magnetic force and F is gravity component (perpendicular to surface)

ILB - mg*sin? = 0 Here B has a horizontal component which is given by B = B*cos ? = 0.1*cos 20 = 0.09397

vLB/R * LB - mg*sin? = 0

vL2B2/R - mg*sin? = 0

now, isolate v and solve for v, we get

v = mg*sin? * R / L2B2

v = 1e-4*9.81*sin20*152 / 0.22 0.93972

v = 144.386 m/s

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