
#1
Apr305, 11:16 PM

P: 26

Suppose the light bulb in Figure 22.4b is a 60.0W bulb with a resistance of 240 W. The magnetic field has a magnitude of 0.40 T, and the length of the rod is 0.60 m. The only resistance in the circuit is that due to the bulb. Minimally, how long would the rails on which the moving rod slides have to be, in order that the bulb can remain lit for onehalf second?
Anyone know how i can start going about solving this :)? 



#2
Apr405, 12:17 AM

P: 40

Faraday's law states that the induced emf is equal to the negative rate of change of magnetic flux. Also the rate of change of magnetic flux is equal to B*dA/dt, since the magnetic field is completely perpendicular to the entire area.
Emf = B*dA/dt, you can relate this to the information given in the problem remember that emf = I*R 



#3
Apr405, 08:57 PM

P: 26

please help me..
i dont know much calculus P = V^2/R im assuming V = emf in the equation emf = vBL so 60W = emf^2/240ohms emf = 120V.. but they do not give me a velocity or anything im stumped :/ 



#4
Apr405, 10:15 PM

P: 40

Magnetic fields inducing electric fields  turning on lightbulb with magnetic field
The using that V = vBL you can solve for velocity. You already calculated V and B and L are given.
Assuming the velocity is constant you can just multiply 1/2 a second to find the rail length. 


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