A conducting rod of length l and mass m is free to slide on a pair of vertical r

Question

A conducting rod of length l and mass m is free to slide on a pair of vertical rails which are connected at one end by resistance R. Ignore friction and resistance in the rails and the rod. A uniform field is applied perpendicular to the plane of the rod and the rails. R = 10 ohms, l = 50cm; m = 5g, B = 0.8T into the paper. If the rod is released from rest, it will approach a constant terminal verlocity, v.

A. ther terminal velocity, v is? (the answer is 3.06 m/s i need to see the work of how it is done)

B. When the rod is falling with this terminal velocisy, what is the current in the rod? (the answer is 0.12 A but i need to see the work of how it is done)

Explanation / Answer

a)

the weight of the rod is equal to magnetic force on the rod

F = mg = IBL

rewrite equation for I

I = mg/BL

From the Ohm's the potential diffeerence across the bar is

V = IR = (mg/BL)* R

Terminal velocity can expressed as

VT =  V/(BL)

    = mgR/(BL)2

    = (0.005 kg) ( 9.8 m/s^2)( 10 ohm)/ (( 0.8 T) ( 0.5 m))^2

     =3.06m/s
(b) current can be expressed as

I=mg/BL

=(0.005 kg) ( 9.8 m/s^2)/( 0.8 T) ( 0.5 m)

=0.12 A

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