What is the terminal (steady) velocity of the bar as it slides frictionlessly do

Question

What is the terminal (steady) velocity of the bar as it slides frictionlessly down the rails?

Two resistanceless rails rest 20 cm apart on a 4.0 degree ramp. They are joined at the bottom by a 0.75 Ohm resistor. At the top a copper bar of mass 0.075 kg (ignore its resistance) is laid across the rails. The whole apparatus is immersed in a vertical 0.75 T field. What is the terminal (steady) velocity of the bar as it slides frictionlessly down the rails? Express your answer using two significant figures.

Explanation / Answer

The component of the velocity of the bar is vcos ,

the induced emf is
      E = BLvcos
the current in the wire
       I = E /R

= (BLvcos )/R

into the page.

       FB = ILB= (B2L2vcos )/R

For the wire to slide down at a steady speed, the net force must be zero. If we consider the components along the rail, we have
       FBcos - mgsin = 0,
       [(B2L2vcos )/R]cos = B2Lvcos2)/R = mgsin

       (0.75 T)2(0.20 m)2v(cos2 4°)/(0.75 ) = 0.075 kg)(9.8 m/s2)sin4°,

v= 0.075 kg)(9.8 m/s2)sin4°/ (0.75 T)2(0.20 m)2(cos2 4°)/(0.75 )
v = 0.26 m/s

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