Using Faraday\'s Law, calculate the induced emf around the loop in the figure th

Question

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.

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

induced emf is e = B*l*v = 0.5*0.32*5.8 = 0.928 V

upper end of the rod is positive and lower end of the rod is negative
magnetic force acting on the wire should be equal to applied force

F_magnetic = F_applied

B*i*L = 5.2

0.5*i*0.32 = 5.2

current is i = 32.5 A

current is negative here for counter clockwise direction

Resistance of the loop is R = e/i = 0.928/32.5 = 0.02855 ohm

Power dissipated is P = i^2*R = 32.5^2*0.02855 = 30.16 W

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

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