Consider the circuit shown in the following figure Figure 1
Consider the circuit shown in the following figure (Figure 1) , but with the bar moving to the right with speed v. The bar has length 0.367 m , R = 45.3 , and B =0.650 T
1.) At an instant when the resistor is dissipating electrical energy at a rate of 0.807 J/s , what is the speed of the bar?
Solution
0.922 J/s = 0.922W
P = V² / R
V = ( P R ) = ( 0.807 W * 45.3 ) = 6.046 V
V_ind = N B v
v = V_ind / ( N B ) = 6.046 / ( 1 * 0.650T * 0.367m ) = 25.34 m/s
N is number of turns. In this case N = 1 because it is a rod
