PHYS 241/spring2014: Assignment EXAM02SP11

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1 PHYS 241/spring2014: Assignment EXAM02SP11 For user = meiles Logout User: meiles sp11ex02q07 [10 points] (Last updated: Mon Apr 11 16:54: ) [meiles] Magnetic field on the axis of solenoid A solenoid 2.9 m long has a radius of 0.85 cm and 560 turns. It carries a current I of 2.5 A. What is the approximate magnetic field B on the axis of the solenoid? a mt b mt c T d mt e mt sp11ex02q08 [10 points] (Last updated: Thu Apr 7 14:37: ) [meiles] Torque on a coil

2 What is the maximum torque on a 1000 turn circular coil of radius 0.75 cm that carries a current of 1.4 ma and resides in a uniform magnetic field of 0.25 T? a N m b N m c N m d N m e N m [10 points] Maximum number of attempts for credit = 5. (Fraction of credit after that: 0.5 up to 10 attempts) sp11ex02q04 [10 points] (Last updated: Thu Mar 31 09:28: ) [meiles] Wire bent in a funny shape When a current is passed through the flexible wire shown in the Figure below, will it tend to bunch up or form a circle?

3 Figure a. Resistivity of the wire is needed to answer this question. b. The wire will tend to form a circle. c. The wire will tend to bunch up. d. The length of the wire is needed to answer this question. e. Current in the wire is needed to answer this question. sp11ex02q09 [10 points] (Last updated: Fri Apr 1 08:35: ) [meiles] Charging a capacitor in a circuit with switch

4 The circuit as shown in the Figure above shows a capacitor C, two ideal batteries, two resistors, and a switch S. Initially S has been open for a long time. If it is then closed for a long time, by how much does the charge on the capacitor change? Assume C = 53 F, 1 = 3.4 V, 2 = 0.52 V, R 1 = 0.1, and R 2 = a C b C c C d C e C sp11ex02q03 [10 points] (Last updated: Thu Apr 7 14:36: ) [meiles] Capacitor with a dielectric A certain dielectric with a dielectric constant = 24 can withstand an electric field of V/m. Suppose we want to use this dielectric to construct a 0.95 µf capacitor that can withstand a potential

5 difference of 1830 V. (a) What is the minimum plate separation? (b) What must the area of the plates be? a. (a) mm; (b) cm 2 b. (a) mm; (b) cm 2 c. (a) mm; (b) 2046 cm 2 d. (a) mm; (b) 4092 cm 2 e. (a) mm; (b) 2046 cm 2 sp11ex02q10 [10 points] (Last updated: Thu Mar 31 10:34: ) [meiles] Combination of electric and magnetic fields An electric field of 1.8 kv/m and a magnetic field of 0.55 T act on a moving electron to produce no net force. Calculate the minimum speed V of the electron. a m/s b m/s c m/s d m/s e m/s sp11ex02q02 [10 points] (Last updated: Thu Mar 31 11:25: ) [meiles] Multi loop circuit

6 For the circuit shown above, = 10 V and R = 3. Find the potential difference between points a and b. State the magnitude. a. 4 V b V c. 6 V d. 8 V e. 12 V HELP: Using symmetry wisely will save time and effort.

7 sp11ex02q11 [10 points] (Last updated: Thu Mar 31 09:29: ) [meiles] LR circuit with a switch and energy source In the circuit shown in the Figure above, = 10 V, R 1 = 4.2, R 2 = 9.5, and L = 4.8 H. For the situation where the switch S just closed, calculate the current i 1 through R 1. a A b A c A d. 0 A e A sp11ex02q01 [10 points] (Last updated: Wed Mar 30 11:25: ) [meiles] Battery with internal resistance

8 In the figure above, the emf is 6 V and R = The rate of Joule heating in R is 8 W. What is the value of r? a b c d e sp11ex02q12 [10 points] (Last updated: Thu Mar 31 09:30: ) [meiles]

9 Magnetic field on the surface of the wire A 10 gauge bare copper wire (2.6mm in diameter) can carry a current of 50 A without overheating. For this current what is the magnetic field at the surface of the wire? a. 0 b mt c mt d mt e mt Logout CHIP ID: meiles

10 1 Magnetic field of a solenoid is In[1]:= Bsol mu0 NN L i Out[1]= In[2]:= i mu0 NN L Bsol. mu0 Pi 4*^-7, NN 560, L 2.9, i 2.5 Out[2]= Torque is defined : In[9]:= torque p Bfield Sin theta Out[9]= Bfield p Sin theta the magnetic moment is : In[11]:= Out[11]= p NN A i A i NN so : In[12]:= torque. NN 1000, A Pi ^ 2, i 1.4*^-3, Bfield 0.25, theta Pi 2 Out[12]= RH rule : magnetic field due to the current at the location of the current on the opposite side points into the page ; then the force points out. Thus the force all points out so it forms a circle. 4 The initial charge is : In[13]:= Qin c E2 Out[13]= c E2 The final charge is more complicated. The current only goes around the outer loop : In[19]:= Out[19]= current E1 E2 R1 R2 E1 E2 R1 R2 Now we can calculate the voltage drop across, say, R1 : In[20]:= vdrop R1 current Out[20]= E1 E2 R1 R1 R2 so, the voltage across the capacitor is equal to the voltage across the other side of the parallel circuit : careful with that minus sign In[24]:= Vcap E1 vdrop Out[24]= In[25]:= E1 E2 R1 E1 R1 R2 Now we can calculate the final charge : Qfi c Vcap Out[25]= c E1 E1 E2 R1 R1 R2 Printed by Mathematica for Students

11 2 examc.nb In[26]:= Qfi Qin. c 53*^-6, E1 3.4, E2 0.52, R1 0.1, R Out[26]= Capacitance definition is : In[27]:= Out[27]= Capp e0 A d A e0 d Electric field is : In[28]:= Efield V d Out[28]= V d So d is : In[31]:= Solve Efield 4*^7, d. V Out[31]= d And then the area is : In[33]:= Out[33]= Solve Capp 0.95*^-6, A. d `, e *^-12 A Don ' t forget to multiply e0 by K 6 Forces balance out : In[34]:= Solve q Ef q v Bf, v Out[34]= v Ef Bf In[35]:= 1.8*^ Out[35]= Homework prob. 8 Inductors skip 9 Let ' s put power in terms of current and resistance : In[43]:= P i ^ 2 R Out[43]= i 2 R Not only that, but, using Ohm ' s law, In[44]:= i e r R e Out[44]= In[45]:= r R Solve P 8, r Out[45]= r e R 4 R, r e R 4 R In[46]:= 1 2 e R 4 R. e 6, R Out[46]= Printed by Mathematica for Students

12 examc.nb 3 10 Magnetic field of a wire : In[47]:= Bfield mu0 II 2 Pi r Out[47]= In[48]:= II mu0 2 Π r Bfield. mu0 Pi 4*^-7, II 50, r Out[48]= Printed by Mathematica for Students

PHY 131 Review Session Fall 2015 PART 1:

PHY 131 Review Session Fall 2015 PART 1: 1. Consider the electric field from a point charge. As you move farther away from the point charge, the electric field decreases at a rate of 1/r 2 with r being