PHYS 241/spring2014: Assignment EXAM02SP13 · 2014. 4. 15. · PHYS 241/spring2014: Assignment...

PHYS 241/spring2014: Assignment EXAM02SP13

User: meilesFor user = meilesLogout

sp13ex02q04 [10 points] (Last updated: Wed Mar 20 14:38:31 2013) [meiles]

Electron paths in a magnetic field 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!

The next three questions pertain to the following situation:

Two electrons labeled A and B possess negative charge q = 1.6 1019 C, and have mass m = 9.1 1031 kg. The electrons are accelerated to different velocities and are sent into a region containing a constant,uniform magnetic field of magnitude 1.5 T but with unknown direction. The electrons' initial velocities arein the +x direction, and they enter the Bfield region at the origin. Within the field region, they followcurved paths confined to the xyplane (z = 0), then emerge at different locations on the yaxis: electron Aemerges at y = a, and electron B emerges at y = b, where b > a as shown.

(a) What is the direction of the uniform magnetic field in the region x > 0?

a. +z direction (out of thepage)

b. z direction (into the page)c. +x direction (to the right)d. x direction (to the left)

e. +y direction (upward)

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

(b) We know that the electrons have the same charge and the same mass. What is the ration of theirvelocities vA/vB?

a. vA/vB = b2/a2

b. vA/vB = a2/b2

c. vA/vB = a/bd. vA/vB = b/a


(c) If an electric field of magnitude E = 300 V/m is switched on before the electrons enter the field region,then electron A is observed to move undeflected, in a straight line along the xaxis. The electric field isperpendicular to both the magnetic field and the initial velocity of the electrons, and occupies the sameregion as the magnetic field. What is the speed of electron A?

a. vA = 0 m/sb. vA = 0.005 m/sc. Cannot move in a straight line at any speedd. vA = 200 m/se. vA = 450 m/s



A loop in a mgnetic field

Don't Forget to Give Answers in Specified Units!


the figure!

The next three questions pertain to the following situation: A rectangular wire loop of height h, width w, and net electrical resistance R lies in the xy plane. As shownin the figure above, the loop is fully contained within a region of spatially uniform but timevaryingmagnetic field. A graph of the magnetic field's time dependence is shown below, along with the values ofall parameters.

(a) Calculate the magnitude of the current induced around the wire loop at time t = 5 sec:

a. |I| = 12 mAb. |I| = 18 mAc. |I| = 2.0

mAd. |I| = 9.0

mAe. |I| = 1.3

mA


(b) What is the direction of the current induced around the wire loop at time t = 5 sec?

a. counterclockwiseb. clockwise


(c) Compare the magnitude of the current at t = 5 seconds, I5, to the magnitude of the current at t = 6

seconds, I6:

a. |I5| < |I6|b. |I5| > |I6|c. |I5| = |I6|



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


the figure!

Calculate the magnetic field B at C, the common center of the semicircular arcs AD and HJ in the Figureabove. The two arcs of radii R1 = 0.13 m and R2 = 0.23 m, respectively, form part of the circuit AHJDAcarrying current i = 3.5 A.

a) Find the magnitude of B.

a. 5.52 106 T

b. 3.68 106 Tc. 13.24 106 Td. 7.35 106 Te. 26.48 106 T


b) Choose the correct direction.

a. into the pageb. towards the bottom of the pagec. towards the left of the paged. towards the right of the pagee. out of the page



Windshield wiper by an RC circuit Don't Forget to Give Answers in Specified Units!


the figure!

A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipersare triggered when the voltage across the 125 F capacitor reaches 10.0 V; then the capacitor is quicklydischarged (through a much smaller resistor) and the cycle repeats. What resistance should be used in thecharging circuit if the wipers are to operate once every 1.80 s?

a. 8.26 kb. 8.15 kc. 8.04 kd. 8.19 ke. 8.11 k


sp13ex02q02 [10 points] (Last updated: Mon Mar 18 10:47:20 2013) [meiles]

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


the figure!

An electron that has velocity v = (2.9 e6 m/s)i + (1.1 e6 m/s)j moves through a magnetic field B = (0.355T)i (0.566 T)j.

a) Find the magnitude of the force on the electron.

a. 32.5 1014 Nb. 37 1014 Nc. 27.2 1014 Nd. 48.7 1014 Ne. 65.2 1014 N


b) In what direction does the force act?

a. Negative y directionb. Positive z directionc. Positive x directiond. Negative z directione. Negative x direction



Force on an electron due to a current in a wire Don't Forget to Give Answers in Specified Units!


the figure!

A long straight wire carries a current of 40 A to the right. An electron, traveling at 2.7 107 m/s, is 5.3 cmfrom the wire. What force, magnitude and direction, acts on the electron if the electron velocity is directedtoward the wire?

a. 6.52 1016 N, parallel to the current in the wireb. 13.04 1016 N, parallel to the current in the wirec. 13.04 1016 N, antiparallel to the current in the wired. 6.52 1016 N, perpendicular to the current and away from the wiree. 6.52 1016 N, antiparallel to the current in the wire


sp13ex02q01 [10 points] (Last updated: Tue Mar 19 08:50:09 2013) [meiles]

Equivalent resistance 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!

Each of the resistors in the diagram above has a resistance of 12 . The resistance of the entire circuit is

a. None of the other answers is correctb. 25 c. 5.76 d. 48 e. 120



Concentric conductors Don't Forget to Give Answers in Specified Units!


the figure!

A long, thin wire carrying constant current I = 2 A into the page is surrounded by a concentric cylindricalhollow wire of inner radius a = 0.12 m, and outer radius b = 0.26 m, carrying total current I = 4 A directedout of the page, as shown. Assume the current in the cylindrical hollow wire is distributed uniformly overits crosssectional area. At what radius is B = 0 in the region a < r < b inside the hollow wire?

a. 0.20 mb. 0.25 mc. 0.19 md. 0.21 me. The magnetic field is not zero anywhere inside the hollow wire



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


the figure!

What is the voltage, V3, across R3 if V = 6 Volts, R1 = R4 = 10 , and R2 = R3 = 5 , and

a. 2.64 Voltsb. 3.64 Voltsc. 5.36 Voltsd. 4.36 Voltse. 1.64 Volts


Logout

CHIP ID: meiles

1 aL Right hand rule works if B points into the page ,

but since they ' re electrons, must be negative , so out of the page .

bL Relate the lorentz force to centripetal acceleration to solve for the radius as a function of v :

In[8]:= Solve @q * v * B m * v ^ 2 r , r D

Out[8]= :: r ®m v

B q>>

So v and r are proportional, so vA vB = A B

c L The electric force must balance the loretnz magnetic force :

In[9]:= Solve @q * efield q * v * B, v D

Out[9]= :: v ®efield

B>>

In[10]:= 300 1.5

Out[10]= 200.

2L Skip !

3L

aL Magnetic field of a semicircle is :

In[11]:= Bcenter = mu0 * i 4 r

Out[11]=i mu0

4 r

RH rule states that the magentic field from the close semicircle is into the screen ,

magnetic field from the far one is out of the screen . So :

In[12]:= Bin = - Bcenter . 8mu0 ® Pi * 4*^-7, i ® 3.5, r ® 0.13<;

Bout = Bcenter . 8mu0 -> Pi * 4*^-7, i ® 3.5, r ® 0.23<;

B = Bin + Bout

Out[13]= - 3.67745 ´ 10- 6

4L The voltage across the capacitor is :

In[38]:= Vc = Q Capp

Out[38]=Q

Capp

Q is given by the charging formula :

In[39]:= Q = Capp * Vemf * H 1 - Exp@-t R CappDL

Out[39]= Capp 1 - ã-

t

Capp R Vemf

So :

In[40]:= Solve @10 Vc , R D

Solve :: ifun : Inverse functions are being used by Solve , sosome solutions may not be found ; use Reduce for complete solution information .

Out[40]= :: R ® -t

Capp LogB-10 - Vemf

VemfF

>>

Printed by Mathematica for Students

In[41]:= -

t

Capp Log B-10 - Vemf

VemfF

. 8t ® 1.8, Qf ® 125 ^ - 6 * 12, Vemf ® 12, Capp ® 125*^-6<

Out[41]= 8036.79

5L Magnitude :

magnitude of v is :

In[47]:= vmag = [email protected] ^ 2 + 1.1 ^ 2D

Out[47]= 3.10161

magnitude of B is

In[48]:= Bmag = [email protected] ^ 2 + 0.566 ^ 2D

Out[48]= 0.668118

the angle between vectors is :

In[56]:= theta = 180 Pi * ArcTan @1.1 2.9D - 180 Pi * ArcTan @-0.566 0.355D

Out[56]= 78.676

So the mag of force is :

In[57]:= 1.602*^-19 * vmag * 1*^6 * Bmag * Sin @theta * Pi 180D

Out[57]= 3.2551 ´ 10- 13

RH rule gives negative z, opposite since it ' s an electron .

6L Magnetic field of wire is

In[59]:= Bwire = mu0 * ii 2 Pi r

Out[59]=ii mu0

2 Π r

so the force is

In[61]:= Fmag = q * v * Bwire . 8q ® 1.602*^-19, v ® 2.7*^7, ii ® 40, mu0 ® Pi * 4*^-7, r ® 0.053<

Out[61]= 6.52891 ´ 10- 16

RH rule gives parallel to wire .

7L

In[62]:= Requiv = H 3 * 1 12L ^ H- 1L + H 2 12L ^ H- 1L + H 4 12L ^ H- 1L + 12

Out[62]= 25

8L Ampere ' s law states that B = 0 if the enclosed current vanishes. So,

we need to surround 2 amps of current in that outer wire . The current density is

In[68]:= Idens = I0 H Pi * b ^ 2 - Pi * a ^ 2L

Out[68]=I0

- a 2 Π + b 2 Π

So,

2 ExamE.nb


In[69]:= Solve @2 Idens * H Pi * r ^ 2 - Pi * a ^ 2L, r D

Out[69]= :: r ® -- 2 a 4 + 4 a 2 b 2 - 2 b 4 + a 4 I0 - a 2 b 2 I0

a 2 I0 - b 2 I0

>,

: r ®- 2 a 4 + 4 a 2 b 2 - 2 b 4 + a 4 I0 - a 2 b 2 I0

a 2 I0 - b 2 I0

>>

In[70]:=- 2 a 4

+ 4 a 2 b 2- 2 b 4

+ a 4 I0 - a 2 b 2 I0

a 2 I0 - b 2 I0

. 8a ® 0.12, b ® 0.26, I0 ® 4<

Out[70]= 0.202485 + 0. ä

9L Set up loop equations :

In[74]:= eq1 = Vv - I1 * R1 - R3 * H I1 - I2L 0;

eq2 = - I2 * R2 - I2 * R4 - R3 * H I2 - I1L 0;

In[76]:= Solve @8eq1, eq2<, 8I1, I2<D

Out[76]= ::I1 ®HR2 + R3 + R4L Vv

R1 R2 + R1 R3 + R2 R3 + R1 R4 + R3 R4, I2 ®

R3 Vv

R1 R2 + R1 R3 + R2 R3 + R1 R4 + R3 R4>>

In[77]:= I1 =

H R2 + R3 + R4L Vv

R1 R2 + R1 R3 + R2 R3 + R1 R4 + R3 R4

Out[77]=HR2 + R3 + R4L Vv

R1 R2 + R1 R3 + R2 R3 + R1 R4 + R3 R4

In[78]:= I2 =

R3 Vv

R1 R2 + R1 R3 + R2 R3 + R1 R4 + R3 R4

Out[78]=R3 Vv

R1 R2 + R1 R3 + R2 R3 + R1 R4 + R3 R4

So the voltage is :

In[80]:= V3 = R3 * H I1 - I2L . 8R1 ® 10, R4 ® 10, R2 ® 5, R3 ® 5, Vv ® 6.<

Out[80]= 1.63636

ExamE.nb 3


PHYS 241/spring2014: Assignment EXAM02SP13 · 2014. 4. 15. · PHYS 241/spring2014: Assignment...

Documents

Transcript of PHYS 241/spring2014: Assignment EXAM02SP13 · 2014. 4. 15. · PHYS 241/spring2014: Assignment...