There are two long parallel rails D=10cm apart. The are connected by a battery o
ID: 2127346 • Letter: T
Question
There are two long parallel rails D=10cm apart. The are connected by a battery on the left side supplying a potential difference of 100V, positive on the upper rail and negative on the lower rail. There is a metal rail with a resistance of 10 ohms across the bar, it can slide to the left or right. Friction is negligible. There is a uniform magnetic field of 1.0T.
1. In what direction should the magnetic field be to give maximum acceleration to the right.
2. In an instant if the current in the circuit is 8A and the bar is travelling 200 m/s to the right with what rate does the magnetic field do work on the bar.
3. Compare this to the power from the battery, is there conservation of energy.
4. With a diagram show emf arising identifying the force doing the work. What expression shows how the induced emf is dependent on the velocity of the bar.
5. Explain why if the velocity is 200 m/s the current is 8A rather than 10A from I=v/r.
6. What is the maximum velocity the bar can achieve from this 100V. If the magnetic field was stronger would the velocity be smaller or greater and why.
Explanation / Answer
1) B = into the plane
2) F = magnetic force = i*L*B = 8*0.1*1 = 0.8 N
rate of work done = F*velocity = 0.8*200 = 160 w
3) power supply by battery =100* 8 = 800 w
4) induce emf = v*B*L = 200*1*0.1 = 20 V
5) due to induced emf , voltage across bar = 20 v
current = (battery voltage- induced emf)/resistance = (100-20)/10 = 8 A
6) v = maximum velocity = 100/(B*L) = 100/(1*0.1) = 1000 m/s
if magnetic fild is stronger then Velocity is smaller , because velocity is inversly propotional to magnetic filed