Question

A solenoid made of magnet wire (#20 gauge) with 12.3 turns per cm is attached to a 536 Ω resistor and a 110 V DC source. The diameter of the solenoid is 1.8 cm and its length is 15 cm a. (5 points) What is the self-inductance of the solenoid? b. (5 points) The emf induced in the solenoid depends on a changing current. After a long time the current will stop changing, and the only voltage drop in the circuit will be across the resistor. What will the current be in the resistor? (5 points) At this time, what is the flux through the solenoid. (Hint: treat the magnetic field as constant inside the coil) C. d. (5 points) At this time, how much energy is stored in the solenoid's magnetic field? If we attach an aluminium rod to conducting rails, we can induce an emf in the circuit by moving it through a magnetic field. If we orient the rails vertically, eventually the rod will reach a terminal velocity. Assume that you can ignore the resistance of the rails. The rod is 30 cm long, has a mass of 130 g, and a resistance of 50 uS2 (5 points) If the rod moves through a 500 Gs magnetic field, calculate the current in the circuit. (In reality, the rails would have much more resistance than the rod. This would a. substantially limit the current, but we won't worry about that) b. (10 points) Calculate the velocity that the rod falls at needed to produce this current We want to go perform the same experiment that Faraday did with the Thames river at the Schuylkill river, instead. At each end of the Market street bridge, we lower two copper plates into the water and then connect them to a voltmeter, which reads 7.03 mV (10 points) The bridge is about 100 meters across, and we measure the river to be flowing at about 1.5 meters per second. What is component of Earth's magnetic field perpendicular to the river? (5 points) The river flows southward. Which side of the river has the higher voltage: east or west? Justify your answer. a. b.

Answer 1

Self inductance_(L) = mu_0 n^2 (Al) n rightarrow number of turns per unit length A rightarrow area of cross section l rightarrow length of n = (12.3) (100) = 1230 term/m a = pi r^2, r = 0.9 cm = (pi) (0.9) times 106 - 2)^2 l = 0. 15 L = mu_0 n^2 Al L = 7.26 times 106 - 5 H I = epsilon/R = 110/536 A I = 0.2 A phi = flux = (N) (B) (A) B = mu_0 n I N rightarrow number of turns n rightarrow number of turns per unit length n = 1230 turns/m I = 0.2 A l = 0.15 m N/l = n n = (nl) phi = N B A = (nl) (mu_0 n I) A = (mu_0) (n^2) (Al) (i) phi = 1.5 times 10^-5 T - m^2 Energy = 1/2 L I^2 Energy = 1.53 times 10^-6 Joule L = 7.26