A conducting rod is pulled horizontally with constant force F= 4.60 N along a se

Question

A conducting rod is pulled horizontally with constant force F= 4.60 N along a set of rails separated by d= 0.320 m. A uniform magnetic field B= 0.500 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 4.70 m/s.

1. Using Faraday's Law, calculate the induced emf around the loop in the figure that is caused by the changing flux. Assign clockwise to be the positive direction for emf.

2. The emf around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction.)

3. From your previous results, what must be the electrical resistance of the loop? (The resistance of the rails is negligible compared to the resistance of the rod, so the resistance of the loop is constant.)

4. The rate at which the external force does mechanical work must be equal to the rate at which energy is dissipated in the circuit. What is that rate of energy dissipation (power dissipated)?

Explanation / Answer

F= 4.60 N ; d= 0.320 m.; B= 0.500 T ; v= 4.70 m/s.

1) Induced EMF is given by the equation E=B*d*v

where B -> Magnetic field strength

d-> Length of the road

v -> velocity of the road

E= .5 * .32 * 4.7

= 0.752 V

2) we know force acting F=I*d*B where I is the current through it

I = F/(d*B) = 4.6/(.32*.5) = 28.75 A

Using flemings left hand rule, direction of current will be in clockwise direction

3) Resistance R=V/I = .752/28.75 = 0.026 ohms

4) Power P= F*v = 4.6* 4.7 = 21.62 W

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