A conducting rod is pulled horizontally with constant force F= 4. 50 N along a s

Question

A conducting rod is pulled horizontally with constant force F= 4. 50 N along a set of rails separated by d= 0. 220 m. A uniform magnetic field B= 0. 700 T is directed into the cage, There is no friction between the rod and the rails, and the rod moves with constant velocity v=3. 70 m/s. 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. When the magnetic field is uniform and normal to the plane of the loop, then the flux is of the product of the field and the area. In this problem it is the area that changes with time, not the field. The emf around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction.) 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.)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

a) Induced emf = B*v*d

= 0.7*3.7*0.22

= 0.5698 volts

= -0.5698 volts

b) in the equilibrium, power requred to move the rod = power dissipated

F*v = emf*I

==> I = F*v/induced emf

= 4.5*3.7/0.5698

= 29.2 A

= -29.2 A(Counter clockwise)

c) resitanse, R = induced emf/I

= 0.5698/29.2

= 0.019 ohms

d) rate of energy dissipated, p = F*v

= 4.5*3.7

= 16.65 Watts

a) Induced emf = B*v*d

= 0.7*3.7*0.22

= 0.5698 volts

= -0.5698 volts

b) in the equilibrium, power requred to move the rod = power dissipated

F*v = emf*I

==> I = F*v/induced emf

= 4.5*3.7/0.5698

= 29.2 A

= -29.2 A(Counter clockwise)

c) resitanse, R = induced emf/I

= 0.5698/29.2

= 0.019 ohms

d) rate of energy dissipated, p = F*v

= 4.5*3.7

= 16.65 Watts

Let me know if it is wrong. I will make corrections.

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