Phys 202 1st Edition Lecture 19Outline of Last Lecture I. RL and RC circuitsOutline of Current Lecture II. Example problemsCurrent LectureTrue/False Example Problem:The figures below show two different situations where a current may be induced in a loop according to Faraday's Law, with the direction given by Lenz' Law. The magnetic field is shown by the x's in Fig. 2. Select true or false for the current in the loop.1. In figure 1, the magnet moves east, and induces current a.Answer: True (use right hand rule, thumb faces the opposite direction of the movement of the magnet. Because the current needs to oppose the decreasing flux.)2. In figure 1, the magnet moves west and induces the current a. Answer: False (induces current B for same reason as stated above) 3. In figure 1, the loop moves west which induces a current a.Answer: True4. In figure 2, the loop moves north and induces a current b. These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Answer: False (In the loop moves north or south then the flux isn’t changing)5. In figure 2, the loop moves east and induces a current b. Answer: True6. In figure 2, the loop moves south and induces a current a. Answer: False (once again, the flux isn’t changing)Solenoid Example Problem:A very long solenoid with a circular cross section and radius r1= 2.70 cm with ns= 300 turns/cm lies inside a short coil of radius r2= 3.80 cm and Nc= 23 turnsA) If the current in the solenoid is ramped at a constant rate from zero to Is= 1.50 A over a time interval of 91.0 ms, what is the magnitude of the emf in the outer coil while the current in the solenoid is changing?B) What is the mutual inductance between the solenoid and the short coil?C) Now reverse the situation. If the current in the short coil is ramped up steadily from zero to Ic= 3.70 A over a time interval of 28.0 ms, what is the magnitude of the emf in the solenoid while the current in the coil is changing?A)If the current is ramped clockwise, then B field flows to the right and flux is changing. EMF will be set up to oppose the change in the flux. Ec is counter clockwise. E=ΔφB/ΔtB=µnsisE=Ncπrs2BE=Ncπrs2µns(Δis/t)=(23)π(.0272)(4πx10-7)(30000)(1.5/.091)Ec=.033VB)Ec=-M(is/t).033=M(1.5/.091)M=.00199HC)Ic: The current is now going from 0A to 3.7A in .028 seconds. Because the coil and solenoid are the same,the mutual inductance between them is the same.
View Full Document
Unlocking...