A conducting rod is pulled horizontally with constant force F= 4.90 N along a se

Question

A conducting rod is pulled horizontally with constant force F= 4.90 N along a set of rails separated by d= 0.520 m. A uniform magnetic field B= 0.600 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 6.70 m/s.

A) 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.

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

C) 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.)

D) 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)?

A conducting rod is pulled horizontally with constant force F= 4.90 N along a set of rails separated by d= 0.520 m. A uniform magnetic field B= 0.600 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 6.70 m/s. A) 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. B) The emf around the loop causes a current to flow. How large is that current? (Again, use a positive value for clockwise direction.) C) 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.) D) 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

d= 0.52 m
v= 6.70 m/s

B= 0.6

=> Rate of change of area = d*v = 3.484 m^2/s
=> Rate of change of magnetic flux = B* (d*v) = 2.0904 Weber/s
(a)

Since Flux is increasing (into the page)
Lenz' Law It tries to oppose the change so, using right hand thumb rule current must be produced in such a way to that B must be (from the page) induced current and hence EMF is in anti clock wise direction
=> its +2.0904 V


(b)

Now since Force * Velocity = Power

We have P =6.70 * 6.70

But P also equals V * I

=> 6.70 * 6.70 = 2.0904* I

=> I = 21.47 Amps


(c)

Now R = V/I

=> R = 2.0904/21.474

=> R = 0.09734 Ohms


(d)

Power is same again

=> P= 6.70*6.70

=> P = 44.89 Watts

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.