Two parallel conducting rails, separated by distance L, are partially laid insid

Question

Two parallel conducting rails, separated by distance L, are partially laid inside uniform magnetic field B perpendicular to the rails (see figure). One end of the parallel rails (the end sticking outside the patch of magnetic field) are connected by a light bulb. The resistance of the light bulb is R. A conducting rod lying on the two rails slides into the field patch at initial speed v0. In the following, assume that the resistances of the rails and the rod are negligible.

7. Find the power dissipated by the bulb as a function of tine P(t). Use the formula P = IE 8. Integrate the power above to find the net energy that has dissipated from the bulb by the time rod comes to a halt and the light put out. (The answer should be no surprise!) 7.1 Two parallel conducting rails, separated by distance L, are partially laid inside uniform magnetic field B perpendicular to the rails (see figure). One end of the parallel rails (the end sticking outside the patch of magnetic field) are connected by a light bulb. The resistance of the light bulb is R. A conducting rod lying on the two rails slides into the field patch at initial speed vo. In the following, assume that the resistances of the rails and the rod are negligible B(O 1. Find the emf induced in the rod when the rod is moving at speed v 2. Find the current running through the bulb when the rod is moving at speed v 3. Find the magnetic force that acts on the rod when the rod is moving at speed v 4. Set up a differential equation for v (Newton's second law). (Hint: a = dt.) 5. Solve the differential equation above and find v asa function of timet. Assume that the rod enters the field patch at t = 0 (with speed to) 6. How far does the rod slide inside the field patch before it eventually comes to a halt?

Explanation / Answer

1) E = vBL

2) I = vBL/R

3) F= evB

4) F= evB = mdv/dt

5) dv/v = eBdt

ln(v) = eBt

v = V0 e^(eBt)

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