The figure shows a zero-resistance rod sliding to the right on two zero-resistan

Question

The figure shows a zero-resistance rod sliding to the right on two zero-resistance rails separated by the distance L = 0.33 m. The rails are connected by a 14.1- resistor, and the entire system is in a uniform magnetic field with a magnitude of 0.750 T.

(a) Find the force that must be exerted on the rod to maintain a constant current of 0.125 A in the resistor.
  30.93 mN

(b) What is the rate of energy dissipation in the resistor?
  220 mW

(c) What is the mechanical power delivered to the rod?
mW

I know the formula is

b^2I^2v^2/R but I don't know how to derive velocity...

Explanation / Answer

F = I*L**B = 30.93 m N

B) P=I^2 x R = 220.03125 mW

c) P= F x velocity = 30.93 *10-3 N * 7.1212 = 220.2590 Mw

you can right it directly from abovr result bcz mechanical power delivered = burned power by resister

we know that

emf = V * B*L ( in moving rod in B field )

emf = I*R = 0.125*14.1 = 1.7625

so 1.7625 =  V * B*L

V= 1.7625 / B*L = 7.1212 m/s

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