Question

Two parallel metal rods form a plane inclined at 20.0° to the horizontal. The upper end of the rods are connected together by a resistor of 146.0 ?. The lower end of the rods are held in an electrically insulating 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/s2.

Answer 1

According to the given problem,

Vector analysis shows that the component of B
perpendicular to the inclined plane is: B' = Bcos?

At terminal velocity F_net = 0: The magnetic force acting
on the block due to increasing induced current in the block
acts in opposition to a constant gravitational force. Thus, the
block reaches its maximum velocity when said forces cancel.

Emf = v*L*B'
i = v*L*B' / R
?F = Fb + Fg = 0
?F = iLB' - mgsin? = 0
?F = [v*L*B' / R]*LB' - mgsin? = 0
Solving for v, terminal velocity:
v = m*g*sin?*R / (L²*B'²)
v = 138.688970254477 m/s
v = 139 m/s .

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