Exam 2 Solutions - Department of Physics

[Pages:11]PHY2049 Fall 2015 ? Acosta, Woodard

Exam 2 Solutions

Note that there are several variations of some problems, indicated by choices in parentheses.

Problem 1: Two light bulbs have resistances of 100 and 300 . They are connected in parallel across a 120V line. What is the total power dissipated by the two bulbs (or the 100 or 300 bulbs)?

This problem was similar to Exercise 26-C done in class.

P = IE = E2 R-1 =

E2 ( R1-1 + R2-1)

=

(120)2

(

1 100

+

1 300

)W

=

192W

;

or 144 W and 48 W individually.

(1.1)

Problem 2: In the circuit shown in the figure both batteries have insignificant internal resistance and the idealized ammeter reads I1 = 4.0A in the direction shown. Find the E.M.F. of the battery (a negative answer

indicates that the E.M.F. polarity is opposite to what is shown).

This problem was similar to Exercise 26-D done in class.

75.0V = E1 = R1I1 + R2 (I1 - I2 ); - E = R3I2 + R2 (I2 - I1); I1 = 4.0A; R1 = 12.0; R2 = 48.0; R3 = 15.0; (1.2)

We note that the

first

equation in (1.2) implies

I2

=

1 R2

((R1 + R2 )I1 - E1) , which

can

be immediately

put

into the

2nd equation to yield E, as,

E = R2I1 - (R2 + R3 )I2

=

R2 I1

-

R2 +R3 R2

((R1

+

R2

)I1

-

E1 )

=

-24.5625V

;

(1.3)

Problem 3: A capacitor with an initial potential difference of V (0) = 150V is discharged through a resistor when a switch between them is closed at t = 0 . At t = 10.0 , the potential difference across the capacitor is V1 = 1.5V . What is the potential difference across the capacitor at t = 20s ?

This is based on homework problem 27.64

PHY2049 Fall 2015 ? Acosta, Woodard Recalling that q(t) = CV (t) = q0 + q1e-t/ in which the discharging-conditions q(0) = CE = CV (0) and q() = 0

give q0 = 0 and q1 = CV (0), we have V (t) = V (0)e-t/ . The third condition V (10.0s) = 1.5V = (150V )e-(10.0s)/

determines

to

be

= 10.0s

/

(- ln

) 1.5

150

=

2.1715s ,

which

determines

the

voltage

at

all

times,

and

so

we

calculate directly V (20.0s) = (150V )e-(20.0s)/(2.1715s) = 0.015V .

Problem 4: In the figure E = 14V , R1 = R3 = 1.00, and R2 = 2. What is the potential difference VA -VB ?

(1.4)

This is based on a Ch.27 homework problem (27.35), and Exercise 26-F done in class

Let the current through the EMF be i, that through R1 be i1, and the current through R2 be i2.

Junction rules: i = i1 + i2 i1 = i2 + i3 i3 = i1 - i2

Left loop rule, substituting R1 = 1 and R2 = 2 : 14 - i1 - 2i2 = 0 i1 = 14 - 2i2

Zig-zag loop rule: 14 - i1 - i3 - i1 = 0

14 - 2i1 - (i1 - i2 ) = 0

14 - 3i1 + i2 = 0

14 - 3(14 - 2i2 ) + i2 = 0

-28 + 7i2 = 0 i2 = 4A i1 = 14 - 2i2 = 6A i = i1 + i2 = 10A

Now VA - VB = i1R1 = (6A)(1) = 6V

PHY2049 Fall 2015 ? Acosta, Woodard Problem 5: A proton travels through uniform

magnetic

and

electric

fields.

The

magnetic

field

is

r B

=

-2.50i^

mT.

At one instant the velocity of the proton is vr = 2000 ^j m/s . At that instant what is the net force acting on the

proton if the electric field is 4.00k^ V/m?

This is problem 28.10, which was assigned in homework

! F

=

q(

! E

+

v!

?

! B)

=

q(

Ek^

+

v^j

?

B(-i^))

=

q(E + vB)k^

(1.5)

= (1.602 ? 10-19 )(4.00 + (2000)(2.50 ? 10-3))k^N = 1.442 ? 10-18 Nk^

r Problem 6: In the figure a charged particle moves into a region of uniform magnetic field B , goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It

r spends 130 ns in the region. What is the magnitude of B ?

This is problem 28.26, which was assigned in homework

(1.6)

We immediately notice that the particle velocity vr = v(- ^j) is deflected in the -i^ direction by the magnetic field

r B

=

Bk^

pointing

out

of

the

page.

Looking

at

the

vector-value

of

the

forces,

! FB

=

FB (-i^)

=

qv!

?

! B

=

q(v(-

^j))

?

( Bk^)

=

qvB(-i^)

qvB

>

0

q

>

0

qm==+mep==11.6.60722??1100-1-927Ck;g;

(1.7)

While in the circular region, the charged particle has constant speed v = vr , and maintains this velocity. This is

because

the magnetic force

r FB

(due to magnetic field

r B

=

Bk^

and velocity

vr

= v dr

= R dr) is

perpendicular to the displacement, and thus does no work. Letting the half-circle have radius R , and letting the

particle have charge q , we have,

dK = dW

=

! FB

?

ds!

=

(qv!

?

! B)

?

Rd!

=

qvBR(r^

?

d!)

=

qvBR(0)

=

0

dK

=

dW

=0

Ki = K f

= Ka ; (1.8)

Hence,

no

kinetic

energy

is

added

or

subtracted

to

the

particle

of

mass

m.

the

velocity

r v

is

of

constant

magnitude. In the circular trajectory, Newton's 2nd Law then is,

F

=

mpa

=

mp

-v2 R

=

-

! FB

mp

-v2 R

=

-qv!

?

! B

mp

-v2 R

=

-qv!

?

! B

mp

v2 R

" =sin 90?=1 = qvB sinv!B! B =

mpv ; (1.9) Rq

In

(1.9),

the

velocity

is

given

by

v

=

2 R T

=

R (1/ 2 )T

,

so,

PHY2049 Fall 2015 ? Acosta, Woodard

v

=

2 R T

=

R

1 2

T

B

=

m ( R p (1/2)T R(+e)

)

=

mp

e

1 2

T

=

(1.672?10-27 kg) (1.602?10-19C)(130?10-9 s)

=

0.252

kg Cs

= 0.252

N

m s

C

= 0.252T

; (1.10)

In the last steps of (1.10), we illustrate the units1 of magnetic field.

Problem 7: In a certain cyclotron a proton moves in a circle of radius 0.5 m. The magnitude of the magnetic field is 1.2 T. What is the kinetic energy of the proton in million electron-volts (MeV)?

This is problem 28-38, which was assigned in homework.

K

=

1 2

mv2

=

1 2

m(C R)2

=

1 2

m(

eB m

R)2

= (1.67?10 kg)( 0.5m) 1

-27

(1.602?10-19 C )(1.2T )

2

1

2

1.67?10-27 kg

1.602?10-13

J MeV

= 17.267MeV ; (1.11)

Problem 8: The figure shows a wood cylinder of mass m = 0.250 kg and length L = 0.100 m, with N = 10 turns of wire wrapped around it longitudinally, so that the plane of the wire contains the long central axis of the

cylinder. The cylinder is released on a plane inclined at an angle to the horizontal, with the plane of the coil

parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude 0.500 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane?

This is problem 28-51, which was assigned in homework.

?

(1.12)

Let the wood-cylinder be of mass m , and have moment of inertia2 I . Newton's 2nd Law for translational

equilibrium between the force of static3 friction f and magnetic force FB and rotational equilibrium between the torque of static friction Rf and magnetic torque B is,

4

4

4

ma = m 0 = 0 = F = f - mg sin; I = 0 = = RiFi sini - fR = RiFi sin 90? - fR = RiFi - Rf ; (1.13)

i =1

i =1

i =1

1 A tesla is a unit of force per unit velocity per unit charge; essentially the units of electric field divided by velocity.

2 The radius of the wooden cylinder is

R , but the wooden-material may be inhomogeneous, so assume

I

1 2

mR2 .

3 CAUTION: The force of static friction is a reaction force, and its magnitude is unknown. The maximum value the force of friction

could take on, if we knew the static-friction-coefficient ?S , is max f = ?S N , where N is a reaction force which has a known contribution mg cos from gravity, but an unknown contribution from the net magnetic force.

PHY2049 Fall 2015 ? Acosta, Woodard

The problem is to find F1, F2, F3, F4 : the magnetic forces upon the four sections of the square-loop shown in the

Figure. The magnitude of the force upon a wire of length l

making angle

with a magnetic field

r B

carrying

current

I

= dq / dt

is given by

F = Il

r B

sin

.

There

are

N

such wires producing identical and superimposing

forces. Thus, let the wood-cylinder be of radius R . Sections 1 and 3 are of length 2R , while Sections 2 and 4

are of length L . Then,

R1F1 = Ri1l 1B sin1N = R(+i)(2R)B sin(90? - )N = +2iR2B cos N;

R3F3 = Ri3l 3B sin3N = R(+i)(2R)B sin(270? - )N = -2iR2B cos N = -R1F1; R2F2 = Ri2l 2B sin2N = (+R)(+iLB sin )N = RiLB sin N; R4F4 = Ri4l 4B sin4N = (-R)(-iLB sin )N = RiLB sin N;

(1.14)

Combining (1.13) and the explicit forces (1.14), and noting the simplification R1F1 = -R3F3, we have,

!##f =m"gs#in #$ 4

Rf = Rmg sin = Ri Fi = R1F1 + R2F2 + R3F3 + R4F4 = RiLBsin N + RiLBsin N + 0 = 2RiLBsin N ; (1.15) i=1

Solving (1.15) for i , we have,

Rmg sin = 2RiLB sin N solveforii = mg =

(0.250kg

)(9.81

m s2

)

=

C 2.453

;

2LBN

2(0.100m)(0.500

(

N m/s

)C

)(10.0)

s

(1.16)

Afterword: We note that the net torque B due to the four branches of the loop has the property,

B =

4

2!RL#=a="are#a=$A/ N Ri Fi = 2RiLBN sin = B 2RL i N sin =

B a i N sin = B

A i sin

B ?% sin =

% B

?

?%

;

(1.17)

i=1

We introduced the magnetic dipole moment vector,

?!

=

! iA

=

iNa!

=

iNan^

,

where

n^ = k^ cos +

^j sin

is the

plane-normal

defining

the

vector-area

r A

=

Nar

=

Nan^

=

N

2LRn^

.

Recall,

also,

that

we

encountered

vector

area

in

our study of the flux that naturally occurred in Gauss's law.

Problem 9: The figure shows, in cross section, two long straight wires held against a plastic cylinder of radius R = 20cm . Wire 1 carries current i1 = 60mA out of the page and is fixed in place at the left side of the cylinder. Wire 2 carries current i2 = 40mA out of the page and can be moved around the cylinder. At what (positive) angle 2 should wire 2 be positioned such that, at the origin, the net magnetic field due to the two currents has magnitude B = 80nT ?

This is problem 29-34, which was worked in class on Oct.14

PHY2049 Fall 2015 ? Acosta, Woodard

(1.18)

The two magnetic

fields

decompose as

r B1

=

B1 ^j

and

r B2

= B2(sin2i^ - cos2 ^j) , so the resultant of

this, using

Amp?re's law to say

B1 =

?0i1 2 R

and

B2

=

?0i2 2 R

(in which we clearly have

i1 =

3 2

i2

),

is,

B

= =

(B2 sin2 )2

+ (B1 - B2 cos2 )2

=

?0i2 2 R

?0i2 2 R

1+

9 4

- 3cos2

2

= cos-1

(1 13

34

sin2 2

+

(

3 2

-

cos2

)2

=

?0i2 2 R

sin2 2

- ( ) ) = cos ( - ( ) ) 2 RB 2 ?0i2

-1 1 13 34

0.2(80?10-2 T ) 2 2(40?10-3 A)

+ =

(

3 2

)2

+

cos2

2

104.4775? ;

-

2

3 2

cos2

(1.19)

Problem 10: In the figure a long straight wire carries a current i1 = 30.0A and a rectangular loop carries current i2 = 20.0A. Take the dimensions to be a = 1.00cm , b = 8.00cm , and L = 30.0cm . In unit vector notation,

what is the force on the loop due to i1 ?

This is problem 29-41, which was assigned in homework.

(1.20) The horizontal wires: Consider two typical wires, a and b , a distance d apart, and carrying respective currents

r ia and ib . A differential element of force dFba acts upon wire- b and is due to wire- a ,

(1.21)

PHY2049 Fall 2015 ? Acosta, Woodard

r

By the Lorentz force law, the differential element of force dFba per unit length dx is due to magnetic field4

r Bb

=

?0ia 2 d

(-

^j) ,

so

we

calculate

the

force

per

unit

length

r

r

f dFba ba dx

as,

( ) ( ) !

fba

=

! dFba dz

=

dqb v!b ? dz

! Ba

=

dqb

dz! dt

?

?0ia 2 d

(-

^j)

dz

=

dqb dt

(k^ dz) ?

dz

?0ia 2 d

(-

^j)

=

ib

?0ia 2 d

k^

?

(-

^j)

=

?0iaib 2 d

i^;

(1.22)

Evidently, the force between wires a and b is in the +i^ direction, and thus is attractive. Using this result

(1.22) upon the two horizontal wires in the Figure (numbered 1 and 3 (note the different coordinates!)),

!

F1

=

! dF1

=

L 0

! dF1 dx

dx

=

L 0

?0i1i2 2 a

^j

dx

=

?0i1i2 L 2 a

^j;

! F3

=

L 0

! dF3 dx

dx

=

L 0

?0i1i2 2 (a + b)

(-

^j)

dx

=

-

?0i1i2 L 2 (a + b)

^j;

(1.23)

The vertical wires: The total forces upon wires 2 and 4 due to wire 0 (of infinite length) are,

!

F2

=

! dF2

=

a+b a

! dF2 dy

dy

=

a+b a

dq2

v!2 ? dy

! B0

dy;

! F4

=

a+b a

! dF2 dy

dy

=

a a+b

dq4

v!4 ? dy

! B0

dy;

(1.24)

The integrands in (1.24) (i.e., the forces per unit y -length) are,

! dF2 dy

=

dq2

v!2 ? dy

! B0

=

i2

(-

^j dy) ? dy

?0i1 2 y

(-

k^)

=

?0i1i2 2 y

i^;

! dF4 dy

=

dq4

v!4 ? dy

! B0

=

i2

(+

^j dy) ? dy

?0i1 2 y

(-

k^)

=

- ?0i1i2 2 y

i^; (1.25)

Combining (1.24) and (1.25), we have,

r

F2

=

a+b a

?0i1i2 2 y

i^ dy

=

?0i1i2 2

a+b

i^

a

dy y

=

?0i1i2 2

i^ ln

a

+b ;

a

r F4

=

a+b a

-?0i1i2 2 y

i^ dy

=

-?0i1i2 2

a+b

i^

a

dy y

=

-?0i1i2 2

i^

ln

a

+ a

b

;(1.26)

Looking

at

(1.26),

we

see

r F2

= -Fr4 ,

so

the

superposition

of

these

two

forces

make

no

contribution.

Hence,

! F

=

! F1

+

! F2

+

! F3

+

! F4

=

! F1

+

! F2

+

! F3

-

! F2

=

! F1

+

! F3

=

?0i1i2 L 2 a

^j -

?0i1i2 L 2 (a + b)

^j

=

?0i1i2 L 2

^j

1 a

-

a

1 +

b

=

(4

?

10-7

T m A

)(30.0 A)(20.0 2

A)(30.0cm)

^j

1 1.00cm

-

1.00cm

1 +

8.00cm

=

0.0032N ^j ;

(1.27)

Problem

11:

The

current

density

r J

inside

a

long,

solid,

cylindrical

wire

of

radius

a

= 3.1?10-3 m

is

in

the

direction of the central axis, and its magnitude varies linearly with radial distance r from the axis according to

r J

=

r J (r)

=

J0 (r

/

a)k^

,

where

J0

=

310

A m2

.

What

is

the

magnitude

of

the

magnetic

field

at

r

/

a

=1/

2

?

You

may

need the Jacobian term r dr d for integration in polar coordinates.

This is problem 29-47, which was assigned in homework.

4 Proof: from Amp?re's law:

r Bwire

?

r dl

=

?0iencl

r Bwire

? r^

ds

=

?0iencl

B wire

2

d

=

?0iencl

Brwire

=

?0iencl 2 d

r^ .

In

this

problem, iencl = ia and (by the right hand rule) r^ = - ^j .

PHY2049 Fall 2015 ? Acosta, Woodard

The current enclosed by an Amp?rian-loop of radius 0 b a is,

iencl = iencl (b) =

diencl

=

di dA = dA wire x.s.

b 2 00

J 0 (r

/ a) dr r d

=

2 J0 a

b

r 2dr

0

=

2 J0 a

1 (b3 3

- 03) =

2 J0b3 3a

;(1.28)

The magnetic field at a distance b from the center of the wire, then, is given from the enclosed current (1.28)

via the law of Amp?re,

" " ! B

?

ds!

=

?0iencl

(b)

! B

?

ds!

=

? 0

2

J

b3

0

3a

B^ ? 2b^ =

? 0

2

J b3 0

solve

for B

3a

B=

? 0

J b2 0 3a

;

(1.29)

From (1.29) follows,

Bb

=

B(

1 2

a)

=

?0

J0

(

1 2

3a

a)2

=

1 12

?0aJ0

=

1 12

(4

?10-7

T m A

)(3.1?10-3

m)(310

A m2

)

=

1.0064 ?10-7 T

;

(1.30)

Problem 12: A solenoid that is L = 95cm long has a radius R = 2.0cm and a winding of N = 1200 turns; it carries a current of i = 3.6A . What is the magnitude of the magnetic field inside the solenoid?

This is problem 29-50, which was assigned in homework.

r The magnetic field B at the center of a solenoid made of a wire carrying current i with n = N / L turns per

meter is of magnitude B = ?0ni .

Therefore,

B=

?0(N / L)i

= 4 ?10-7

T m A

1200 0.95cm

(3.6

A)

=

5.7144mT

.

Problem 13: A wire

loop of lengths

L = 40cm and

W

= 25cm

lies in a magnetic

field

r B

=

(0.08

T ms

)

yt

k^

.

What

are the magnitude and direction of the induced E.M.F.?

This is problem 30-12, which was worked in class on Oct.19

E

=

-

d dt

W

dy

0

L 0

b0

yt

dx

=

-b0

(

1 2

W

2

-

1 2

02

)(L

-

0)

=

-

1 2

b0W

2

L

=

-1.00mV

;

clockwise;

Problem 14: The figure shows a rod of length L = 10cm that is forced to move at constant speed v = 5m / s

along the horizontal rails. The rod, rails and connecting strip at the right form a conducting loop. The rod has a resistance R = 0.4 ; the rest of the loop has negligible resistance. A current i = 100A through the long straight wire at a distance a = 10mm from the loop sets up a nonuniform magnetic field through the loop. At

what rate (in ?W) is thermal energy generated in the rod?

This is problem 30-33, which was worked in class on Oct.21

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