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PHYS 241/spring2014: Assignment EXAM02FALL12

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f12ex02q09 [10 points] (Last updated: Fri Oct 26 13:56:57 2012) [meiles]

Inductors in a series Don't Forget to Give Answers in Specified Units!

If you have a figure in the present problem do not use thenumbers shown in the figures. Please use the numbers given in thetext of the problem unless you are asked to use the numbers from

the figure!

A 3 mH inductor and 6 mH inductor are connected in series and a time varying current is established.When the total emf of the combination is 18 V, the emf (in Volts) of the larger inductor is:

a. 12b. 3c. 4.5d. 6e. 24

____________________ [10 points] Maximum number of attempts for credit = 1.


Inductors in a series Don't Forget to Give Answers in Specified Units!

If you have a figure in the present problem do not use the

numbers shown in the figures. Please use the numbers given in thetext of the problem unless you are asked to use the numbers from

the figure!

An 6.2 mH inductor and 3.1 resistor are wired in series to a 40 V ideal battery. A switch in the circuit isclosed at time 0, at which time the current is zero. After a long time, the current in the resistor and thecurrent in the inductor are:

a. 0, 0b. 6.4 A, 6.4 Ac. 12.9 A, 0d. 12.9 A, 12.9 Ae. 12.9 A, 6.4 A


f12ex02q08 [10 points] (Last updated: Thu Nov 8 18:27:02 2012) [meiles]

Circuit in a magnetic field Don't Forget to Give Answers in Specified Units!


the figure!

The circuit shown is in a uniform magnetic field that is into the page and is decreasing in magnitude at arate of 130 T/s. The current in the circuit (in amperes) is:

a. 0.21

b. 0.58c. 0.18d. 0.77e. 0.4



Inductor and a resistor in series Don't Forget to Give Answers in Specified Units!


the figure!

An 8.0 mH inductor and 2.0 resistor are wired in series to an ideal battery. A switch in the circuit isclosed at time 0, at which time the current is zero. The current reaches half its final value at time:

a. 2.8 msb. 170 sc. 250 sd. 4.0 mse. 3 s


f12ex02q07 [10 points] (Last updated: Tue Oct 30 13:14:57 2012) [meiles]

Rectangular loop Don't Forget to Give Answers in Specified Units!


the figure!

A rectangular loop of wire is placed midway between two long straight parallel conductors as shown. Theconductors carry currents i1 and i2 as indicated. If i1 is increasing and i2 is constant, then the inducedcurrent in the loop is:

a. zerob. counterclockwisec. clockwised. depends on i1 - i2e. depends on i1 +i2



Two parallel wires Don't Forget to Give Answers in Specified Units!


the figure!

Two long straight wires are parallel and carry current in opposite directions. The currents are 7.1 A and15.9 A and the wires are separated by 0.4 cm. The magnetic field (in Telsa) at a point midway between thewires is:

a. 7.1 10-4b. zeroc. 8.8 10-4

d. 23 10-4

e. 15.9 10-4



Magnetic field due to a moving electron Don't Forget to Give Answers in Specified Units!


the figure!

Electrons are going around a circle in a counterclockwise direction as shown. At the center of the circlethey produce a magnetic field that is:

a. to the rightb. out of the

pagec. zerod. into the pagee. to the left


f12ex02q13 [10 points] (Last updated: Mon Oct 29 09:01:26 2012) [meiles]

Charging a capacitor Don't Forget to Give Answers in Specified Units!


the figure!

In the circuit shown, the capacitor is initially uncharged. At time t = 0, switch S is closed. If denotes thetime constant, the approximate current through the 3 resistor when t = /100 is

a. 1 Ab. 3/4

Ac. 1/2

Ad. 3/2

Ae. 3/8

A



Currents in a circuit Don't Forget to Give Answers in Specified Units!

If you have a figure in the present problem do not use the

numbers shown in the figures. Please use the numbers given in thetext of the problem unless you are asked to use the numbers from

the figure!

A portion of a circuit is shown, with the values of the currents given for some branches. What is the valueand direction of the current i?

a. 2 A,down

b. 6 A,down

c. 6 A, upd. 4 A, upe. 4 A,

down


f12ex02q14 [10 points] (Last updated: Wed Nov 7 10:18:39 2012) [meiles]

Magnetic field in a toroid Don't Forget to Give Answers in Specified Units!


the figure!

The tightly wound toroid shown consists of 130 turns of wire, each carrying a current I = 3.1 A. If a = 10cm and b = 16 cm, the magnetic field at r = 8 cm, due to the current in this toroid, is ( 0 = 4 10-7 N/A2)

a. 610 T

b. 500 T

c. zerod. 810

Te. 400

T



EMF in a coil Don't Forget to Give Answers in Specified Units!


the figure!

The magnetic flux through a certain coil is given by

m = (1/50 ) cos( 100 t)

where the units are SI. The coil has 100 turns. The magnitude of the induced EMF (in Volts) when t =1/200 s is

a. 200b. 2/c. 100

d. zeroe. 1/(50 )



Parallel plates in a magnetic field Don't Forget to Give Answers in Specified Units!


the figure!

The apparatus in the figure consists of two parallel plates (shown on edge) and a large magnet (not shown).The field of the magnet is uniform, perpendicular to the electric field between the plates, and directed intothe plane of the paper. The magnitude of the field is 0.42 T. Charged particles with speeds of 5.5 105 m/senter this region through the slit at the left and emerge through the exit slit at the right. What magnitude anddirection must the electric field have so that positively charged particles entering from the left will traverseto the exit slit undeviated?

a. 1.84 106 downb. 1.84 106 V/m upc. 2.31 105 V/m

downd. 2.77 106 V/m

downe. 2.31 105 V/m up



magnetic field due to a moving charge Don't Forget to Give Answers in Specified Units!


the figure!

A positively charged body is moving in the negative z direction as shown. The direction of the magneticfield due to the motion of this charged body at point P is

a. 2b. 4c. 5d. 1e. 3



Magnetic field Don't Forget to Give Answers in Specified Units!


the figure!

A straight conductor carrying current i splits into identical semicircular arcs as shown in the figure, What isthe magnetic field at the center C of the resulting circular loop?

a. zerob. 0i/(2R)

c. 0i/(8R)d. 0i/(2R)e. 0i/R



Proton in a magnetic field Don't Forget to Give Answers in Specified Units!


the figure!

A proton with a charge +e is moving with a speed v at 50 to the direction of a magnetic field B. Thecomponent of the resulting force on the proton in the direction of B is

a. evB sin(50 ) cos(50 )b. Zeroc. evB sin(50 )d. evB tan(50 )e. evB cos(50 )


f12ex02q17 [10 points] (Last updated: Wed Nov 7 10:17:12 2012) [meiles]

Motional EMF Don't Forget to Give Answers in Specified Units!


the figure!

Two straight conducting rails form a right angle where their ends are joined. A conducting bar in contactwith the rails starts at the vertex at time t=0 and moves with a constant velocity of 5.17 m/s along them asshown in the figure. A 0.363 T magnetic field points out of the page. Calculate the emf (in Volts) aroundthe triangle at t = 4 s.

a. 77.6b. 7.5c. 15d. zeroe. 38.8


f12ex02q01 [10 points] (Last updated: Thu Oct 25 09:37:38 2012) [meiles]

Two copper wires Don't Forget to Give Answers in Specified Units!


the figure!

Two copper wires have the same volume, but wire 2 is 21% longer than wire 1. The ratio of the resistancesof the two wires R2/R1 is

a. 1.21b. 1.1c. 0.91d. 0.83e. 1.46


f12ex02q18 [10 points] (Last updated: Tue Oct 30 09:13:16 2012) [meiles]

Power in a resistor Don't Forget to Give Answers in Specified Units!


the figure!

The power dissipated in the unknown resistor (R) is

a. 8 Wb. 6Wc. 4 Wd. 10

We. 2 W


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CHIP ID: meiles

1L SKip !

2L Skip !

3L Skip !

4L Skip !

5L Skip !

6L Mag field is given by

In[1]:= Bline = mu0 * i � 2 � Pi � r

Out[1]=i mu0

2 Π r

The net magnetic field , then , is

In[3]:= B1 = Bline �. 8mu0 ® Pi * 4*^-7, i ® 7.1, r ® 0.004 � 2<

Out[3]= 0.00071

In[4]:= B2 = Bline �. 8mu0 ® Pi * 4*^-7, i ® 15.9, r ® 0.004 � 2<

Out[4]= 0.00159

In[5]:= B1 + B2

Out[5]= 0.0023

Note that we add , not subtract these fields. Check this with the RH rule !

7L Into the page - RH rule but then reverse the direction since it ' s an electron !

8L Current is given by

In[6]:= i = v � req * Exp@-t � tau D

Out[6]=ã

-t

tau v

req

In[10]:= i �. 8v ® 10, req ® 3 + 6, t ® tau � 100.<

Out[10]= 1.10006

This answer is crummy since 1.1 is hardly "approximately " 1 ! But whatever ....

9L Use Kirchoff ' s node law : current in = current out. Let ' s assume i is going in . Then :

In[12]:= Solve @ii + 5 + 4 + 2 � 3 + 2D

Out[12]= 88ii ® -6<<

We got a negative number ! That means it ' s going out, or "down "

10L magnetic field outside of a toroid is zero. r is not inside the toroid . Done .

11L Skip !

12L Coulomb force must balance Lorentz force :

In[13]:= Solve @q * Efield � q * v * Bfield , Efield D

Out[13]= 88Efield ® Bfield v <<

In[14]:= Efield = 5.5*^5 * 0.42

Out[14]= 231 000.

Printed by Mathematica for Students

RH rule gives that the Lorentz force points up. So the electric field must point down .

13L RH Rule ...

14L Zero, by RH rule and symmetry .

15L Trick question ! You ' d like to answer evBsin 50,

since that ' s the magnitude of teh Lorentz force . HOWEVER ,

it ' s asking for the force on the proton IN THE DIRECTION OF B. This

is always zero for magnetic forces as a result of the RH rule � Cross product !

16L Skip !

17L Rememebr that resistance is proportional to length . Don ' t get crazy yet and pick 1.21,

though ! Both have the same volume ,

the problem says. So we need to worry about area as well ! Let ' s re -

write the resistance formula in terms of volume and length :

R = rho * L � A = rho * L ^ 2 � A � L = rho * L ^ 2 � V

Now , when we do the ratio, rho and V cancels and so the ratio is in fact

In[15]:= 1.21 ^ 2

Out[15]= 1.4641

Tricky !

18L We know the current, we can get R from that :

In[16]:= Solve @6 � 2 * H R + 1 � H 1 � 2 + 1 � 4 + 1 � 4LL, R D

Out[16]= 88R ® 2<<

Power is IV = I ^ 2 R

In[17]:= 2 ^ 2 * 2

Out[17]= 8

2 examB.nb

Printed by Mathematica for Students

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For - WeeklyJoys · PDF filenumbers shown in the figures. Please use the numbers given in the text of the problem unless you are asked to use the numbers from