Physics 1c Practical, Spring 2015 Hw 3 Solutions

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1 Physics 1c Practical, Spring 2015 Hw 3 Solutions April 16, Serway (5 points) By Lenz s Law, current will flow in each of the two sub-loops to oppose the change in flux. One can easily see then, that the left sub-loop will have counterclockwise current and the right sub-loop will have clockwise current. Now let s treat the two sub-loops individually We know that " = d. Let A L be the area of the magnetic field circle inside the left sub-loop, and A R be the area of the magnetic field inside the right sub-loop. db " 21 = A L db " 23 = A R (1) (2) Also, we know that I 6 + I 5 = I 3 (where the subscript is the resistance of the corresponding resistor in Ohms) since the left sub-loop is counterclockwise and the right one is clockwise. So, we must solve the following system of equations: 8 >< I 6 R 6 + I 3 R 3 = r1 2 db I 5 R 5 + I 3 R 3 = r2 >: 2 db (3) I 6 + I 5 = I 3 Plugging in =0.1 m,r 2 =0.15 m, db = 100 T/s, R n = n, we obtain by solving this system: I 6 =0.062 A (4) I 3 =0.923 A (5) I 5 =0.860 A (6) where I 3 is up through the resistor, the I 6 and I 5 are down through their respective resistors.

2 Ph1cP Hw 1 Solutions Spring Serway 32.8 (5 points) We know that the self-inductance for an air-core solenoid is: L = µ 0N 2 A l So, plugging in N = 580, l =0.36 m, A = ( 0.08 m 2 ) 2, we obtain: (7) L =5.90 mh (8) The magnitude of the self-induced emf is given by " = L I, and we are given that I =4.00 A/s, so: " = 23.6 mv (9)

3 Ph1cP Hw 1 Solutions Spring Serway (5 points) (3 points) Let R be the 450 resistor. The time constant for an RL circuit is = L R tot. Our e ective value of R tot is given by: R tot =(R 1 + R 0 1 ) 1 (10) So: 1 R + 1 R 0 = L 1 1 R = L R 0 (11) Plugging in our value for R 0 = 450, L = 5 mh, and = 15 µs, we obtain: (2 points) R = 1290 (12) At the moment before the switch is thrown, the circuit is assumed to be in a steady state, so the voltage drop across the inductor will be zero. This means that the voltage drop is entirely across the resistors at this point, suggesting that the total current is I = V R tot = V R + V R. Plugging in for 0 our value of R and R, and V = 24.0 V, we obtain: I = 72.0 ma (13) Since the inductor tries to keep the current flowing even after the switch is closed, the moment the switch is closed the current through the inductor will still be 72.0 ma.

4 Ph1cP Hw 1 Solutions Spring Serway (5 points) Let s start by finding the current through the inductor when the motor gets turned o. Since we are in a steady state, then the voltage drop through the inductor is zero, and thus the voltage drop over the resistor is (12 10) V. By V = IR, we then obtain the current I =2/7.5 A= A. So, right after the motor is turned o, the current in the loop will be A. Now, we want the drop through the armature to be no more than 80 V, so accordingly the voltage over the discharge resistor must be no more than 80 V. So R apple 80 V A = 300. So the maximum resistance is: R max = 300 (14)

5 Ph1cP Hw 1 Solutions Spring QP 5 (5 points) Lamination B is the best lamination. Note that eddy currents will flow to oppose the change in flux, so the eddy currents will be along direction of the loops of wire, wrapping around the bar. Lamination B is perpindicular to the flow for these eddy currents, and the insulation will prevent these eddy currents from dissipating a ton of heat. Lamination A is the worst lamination. As we said in part, the eddy currents wrap around the bar just like the wire loops. Lamination A lets these currents flow pretty much unimpeded since the cross-sections of the lamination are parallel to the loops of eddy currents. This means there will be eddy currents (although the currents are thinner than the distance between lamination planes), and thus plenty of heat dissipation. (c) (0 points) There will be eddy currents in D, since the changing B field will still have components perpindicular to the direction of the plane of lamination (provided the angle is not completely vertical = /2). So in general, the eddy currents will be related to cos where is the angle the lamination plane makes to the horizontal. Of course, the thickness of the eddy currents will still be limited by the thickness of the lamination planes.

6 Ph1cP Hw 1 Solutions Spring QP 6 (3 points) We know that r ~ B = µr ~ H = µ ~ J f for linear ~ B = µ ~ H. So, we can use the cylindrical symmetry to show that: (2 points) We know that the self-inductance is given by the equation is: Z ~B d~a 2 rb = µni B = µni 2 r ~B = µni 2 r ˆ (15) 0 = S = µnid 2 Z r2 = µnid 2 ln r 1 r2 dr r So the total flux is = N 0 = N 2 µid 2 ln r 2. So, since L = /I: or: for L = N 2 L 0. L = N 2 µd 2 ln L 0 = µd 2 ln r2 r2 = LI. Note that for one loop, the flux (16) (17) (18)