A bar of negligible resistance and mass of 17 kg is pulled horizontally across frictionless parallel rails, also of negligible resistance, by a massless string that passes over an ideal pulley and is attached to a suspended mass of 190 g. The uniform magnetic field has a magnitude of 600 mT, and the distance between the rails is 33 cm. The rails are connected at one end by a load resistor of 34 m[tex]\omega [/tex].What is the magnitude of the terminal velocity reached by the bar? Answer in units of m/s.(adsbygoogle = window.adsbygoogle || []).push({});

First I drew a free body diagram on the weight, and found that [tex] F_g= Mg= F_m= ILB [/tex]

so I= Mg/LB

Then I used Ohm's Law.

I= E/R= [tex] 1/R * d\phi /dt [/tex]

and flux= BA

so [tex] d\phi/dt [/tex] = B (dA/dt)= BLV

then I= BLV/R , where V is the terminal velocity

so Mg/LB = BLV/R

Solving for V gives MgR/L^2 *B^2.

Then I plugged in my numbers

V= (.19)(9.8)(.034)/(.33^2)(.6^2)

and I got 1.61 m/s

The second part is: What is the acceleration when the velocity v= 1.1 m/s?

I set 1.1 =(.19)(.034)g/(.33^2)(.6^2) and got 6.68 m/s.. which isn't right.. can someone please help?

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# Terminal velocity problem

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