HW11 Solutions (due Tues, Apr 14) 1. T&M 26.P.024 A 10.0 cm length of wire carries a current of 4.0 A in the positive z direction. The force on this wire due to a magnetic field B is = (-0.2 + 0.2 ) j N. If this wire is rotated so that the current flows in the positive x direction, the force on the wire is F = 0.2 k N. Find the magnetic field vector. Solution:

2. T&M 26.P.028 A 4.5-keV electron (an electron that has a kinetic energy equal to 4.5 keV) moves in a circular orbit that is perpendicular to a magnetic field of 0.325 T. (a) Find the radius of the orbit. Find the (b) frequency and (c) period of the orbital motion. Solution: Picture the Problem (a) We can apply Newton’s 2nd law to the orbiting electron to obtain an expression for the radius of its orbit as a function of its mass m, charge q, speed v, and the magnitude of the magnetic field B. Using the definition of kinetic energy will allow us to express r in terms of m, q, B, and the electron’s kinetic energy K. (b) The period of the orbital motion is given by T = 2πr/v. Substituting for r (or r/v) from Part (a) will eliminate the orbital speed of the

electron and leave us with an expression for T that depends only on m, q, and B. The frequency of the orbital motion is the reciprocal of the period of the orbital motion. (a) Apply Newton’s 2nd law to the orbiting electron to obtain:

r

vmqvB

2

= ⇒qB

mvr =

Express the kinetic energy of the electron:

2

2

1 mvK = ⇒m

Kv

2=

Substituting for v in the expression for r and simplifying yields:

qB

Km

m

K

qB

mr

22==

Substitute numerical values and evaluate r:

( )( )

( )( )mm70.0mm696.0

T325.0C10.6021

eV

J101.602kg10109.9keV5.42

19

1931

==!

""#

$%%&

' !!

= (

((

r

(b, c) Relate the period of the electron’s motion to the radius of its orbit and its orbital speed:

v

rT

!2=

Because qB

mvr = :

qB

m

v

qB

mv

T!

!2

2

==

Substitute numerical values and evaluate T:

( )( )( )

ns 11.0s 10099.1

T 325.0C101.602

kg10109.92

10

19

31

=!=

!

!=

"

"

"#

T

The frequency of the motion is known as the cyclotron frequency and is the reciprocal of the period of the electron’s motion:

GHz1.9ns110.0

11===

Tf

3. T&M 26.P.035 A velocity selector has a magnetic field that has a magnitude equal to 0.28 T and is perpendicular to an electric field that has a magnitude equal to 0.46 MV/m. (a) What must the speed of a particle be for that particle to pass through the velocity selector undeflected? What kinetic energy must (b) protons and (c) electrons have in order to pass through the velocity selector undeflected? Solution: Picture the Problem Suppose that, for positively charged particles, their motion is from left to right through the velocity selector and the electric field is upward. Then the magnetic force must be downward and the magnetic field out of the page. We can apply the condition for translational equilibrium to relate v to E and B. In (b) and (c) we can use the definition of kinetic energy to find the energies of protons and electrons that pass through the velocity selector undeflected. (a) Apply 0=! yF to the particle to obtain:

0magelec =! FF or

0=! qvBqE ⇒B

Ev =

Substitute numerical values and evaluate v:

m/s106.1

m/s1064.1T28.0

MV/m46.0

6

6

!=

!==v

(b) The kinetic energy of protons passing through the velocity selector undeflected is:

( )( )

keV14

J101.602

eV1J1026.2

m/s1064.110673.1

19

15

26kg

27

2

1

2

p2

1p

=

!!!=

!!=

=

"

"

"

vmK

(c) The kinetic energy of electrons passing through the velocity selector undeflected is:

( )( )

eV7.7

J101.602

eV1J1023.1

m/s1064.110109.9

19

18

26kg

31

21

2

e21

e

=

!!!=

!!=

=

"

"

"

vmK

4. T&M 26.P.042 Before entering a mass spectrometer, ions pass through a velocity selector consisting of parallel plates that are separated by 2.0 mm and have a potential difference of 160 V. The magnetic field strength is 0.42 T in the region between the plates. The magnetic field strength in the mass spectrometer is 1.2 T. Find (a) the speed of the ions entering the mass spectrometer and (b) the difference in the diameters of the orbits of singly ionized 238U and 235U. The mass of a 235U ion is 3.903 × 10–25 kg. Solution: Picture the Problem We can apply a condition for equilibrium to ions passing through the velocity selector to obtain an expression relating E, B, and v that we can solve for v. We can, in turn, express E in terms of the potential difference V between the plates of the selector and their separation d. In (b) we can apply Newton’s 2nd law to an ion in the bending field of the spectrometer to relate its diameter to its mass, charge, velocity, and the magnetic field. (a) Apply 0=! yF to the ions in the crossed fields of the velocity selector to obtain:

0magelec =! FF or

0=! qvBqE ⇒B

Ev =

Express the electric field between the plates of the velocity selector in terms of their separation and the potential difference across them:

d

VE =

Substituting for E yields: dB

Vv =

Substitute numerical values and evaluate v:

( )( )

m/s109.1

m/s10905.1T42.0mm2.0

V160

5

5

!=

!==v

(b) Express the difference in the diameters of the orbits of singly ionized 238U and 235U:

235238ddd !=" (1)

Apply cradial

maF =! to an ion in the spectrometer’s magnetic field:

r

vmqvB

2

= ⇒qB

mvr =

Express the diameter of the orbit: qB

mvd

2=

The diameters of the orbits for 238U and 235U are: qB

vmd 238

238

2= and

qB

vmd 235

235

2=

Substitute in equation (1) to obtain:

( )235238

235238

2

22

mmqB

v

qB

vm

qB

vmd

!=

!="

Substitute numerical values and evaluate Δd:

( )( )

( )( )cm1

T2.1C101.602

u

kg106606.1u235u238m/s10905.12

Ä19

275

=!

""#

$%%&

' !(!

= (

(

d

5. T&M 26.P.071 An alpha particle (charge +2e) travels in a circular path of radius 0.50 m in a magnetic field of 0.10 T. Take m = 6.65 10-27 kg for the mass of the alpha particle. (a) Find the period of the alpha particle. (b) Find its speed. (c) Find its kinetic energy. Solution:

6. T&M 26.P.080 A circular loop of wire that has a mass m and a constant current I is in a region with a uniform magnetic field. It is initially in equilibrium and its magnetic moment is aligned with the magnetic field. The loop is given a small angular displacement about an axis through its center and perpendicular to the magnetic field and then released. What is the period of the subsequent motion? (Assume that the only torque exerted on the loop is due to the magnetic field and that there are no other forces acting on the loop. Solution:

7. T&M 26.P.078

A circular loop of wire that has a mass m and a constant current

I is in a region with a uniform magnetic field. It is initially in

equilibrium and its magnetic moment is aligned with the

magnetic field. The loop is given a small angular displacement

about an axis through its center and perpendicular to the

magnetic field and then released. What is the period of the

subsequent motion? (Assume that the only torque exerted on the

loop is due to the magnetic field and that there are no other

forces acting on the loop. Use pi for π, m, I, and B as necessary.)

Solution:

8. T&M 27.P.019 A small current element at the origin has a length of 2.0 mm and carries a current of 2.0 A in the +z direction. Find the magnitude and direction of the magnetic field due to the current element at the point (0, 3.0 m, 4.0 m). Solution: Picture the Problem We can substitute for I and !

"!"

!"d in the Biot-Savart law

(2

0ˆ

4 r

Idd

rB

!=

!"

"

"

µ ), evaluate r and r for (0, 3.0 m, 4.0 m), and substitute to

find .B

!d

The Biot-Savart law for the given current element is given by:

2

0ˆ

4 r

Idd

rB

!=

!"

"

"

µ

Substituting numerical values yields:

( )( )( ) ( )2

2

2

27ˆˆ

mnT400.0ˆˆmm0.2A0.2

N/A100.1rr

drkrk

B!

"=!

!=#

!

Find r and r for the point whose coordinates are (0, 3.0 m, 4.0 m):

( ) ( )kjr ˆm0.4ˆm0.3 +=! ,

m0.5=r , and kjr ˆ

5

4ˆ

5

3ˆ +=

Evaluate B

!d at (0, 3.0 m, 4.0 m):

( ) ( )( )

( )ikjk

B ˆpT6.9m0.5

ˆ5

4ˆ5

3ˆ

mnT400.0m 4.0m,0.3,02

2 !=

"#$

%&'

+()=

!d

9. T&M 27.P.076 Find the magnetic field at point P in the figure below, where I = 15 A and R = 20 cm.

Solution: Picture the Problem Because point P is on the line connecting the straight segments of the conductor, these segments do not contribute to the magnetic field at P. Hence, we can use the expression for the magnetic field at the center of a current loop to find BP. Express the magnetic field at the center of a current loop:

R

IB

2

0µ

=

where R is the radius of the loop.

Express the magnetic field at the center of half a current loop:

R

I

R

IB

422

100

µµ==

Substitute numerical values and evaluate B: B =

4! "10#7 N/A2( ) 15 A( )4 0.20m( )

= 24µT into the page

10. T&M 27.P.081 A long, straight wire carries a current of I1 = 20 A as shown in the figure. A rectangular coil with two sides parallel to the straight wire has sides 5 cm and 10 cm with the near side a distance 2 cm from the wire. The coil carries a current of I2 = 5 A.

Solution: Picture the Problem Let I1 and I2 represent the currents of 20 A and 5.0 A, top

F

!,

sideleft F

!,

bottomF

!, and sideright F

! the forces that act on the horizontal wire, and

1B

!,

2B

!,

3B

!, and

4B

! the magnetic fields at these wire segments due to I1. We’ll need

to take into account the fact that 1B

! and

3B

! are not constant over the segments 1

and 3 of the rectangular coil. Let the +x direction be to the right and the +y direction be upward. Then the +z direction is toward you (i.e., out of the page). Note that only the components of

1B

!,

2B

!,

3B

!, and

4B

! into or out of the page

contribute to the forces acting on the rectangular coil. The +x and +y directions are up the page and to the right. (a) Express the force

1Fd!

acting on a current element !

"dI2

in the top segment of wire:

12top BF

!"!!!= dId

Because ( )i22

!= !!"

dIdI in this segment of the coil and the magnetic field due to I1 is given by

( )kB ˆ

2

10

1!=!

"

"

µ I:

( ) ( )

j

kiF

ˆ2

ˆ2

ˆ

210

10

2top

!

!

!!

"

dII

IdI

!

µ

!

µ

"=

"#"=

Integrate topF

!d to obtain:

j

jF

ˆcm 2.0

cm 0.7ln

2

ˆ2

210

cm 7.0

cm 0.2

210

top

!"#

$%&'=

'= (

)µ

)µ

II

dII

!

!"

Substitute numerical values and evaluate top

F

!:

( )( )( )jjF ˆN 105.2ˆ

cm 2.0

cm 0.7ln

2

A 0.5A 20A

N104

52

7

top

!

!

"!=#$%

&'(

#$%

&'( "

!=)

)!

Express the force

bottomF

!d acting on

a current element !"dI2

in the horizontal segment of wire at the bottom of the coil:

32bottomBF

!"!!!= dId

Because ()i22!!

"dIdI = in this

segment of the coil and the magnetic field due to I1 is given by

( )kB ˆ

2

10

1!=!

"

"

µ I :

( )

j

kiF

ˆ

2

ˆ

2

ˆ

210

10

2bottom

!

!

!!

"

dII

IdId

!

µ

!

µ

=

"#=

Integrate bottomF

!d to obtain:

j

jF

ˆ

cm 2.0

cm 0.7ln

2

ˆ

2

210

cm 7.0

cm 0.2

210

bottom

!"#

$%&

=

= '

(µ

(µ

II

dIId

!

!"

Substitute numerical values and evaluate

bottomF

!:

( )( )( )jjF ˆN 105.2ˆ

cm 2.0

cm 0.7ln

2

A 0.5A 20A

N104

5

2

7

bottom

!

!

"=#$%

&'(

#$%

&'( "

=)

)!

Express the forces

sideleft F

!and

sideright F

!in terms of I2 and

2B

!and

4B

!:

222sideleft BF

!"!!

!= I and

442sideright BF

!"!!

!= I

Express

2B

!and

4B

!:

kB ˆ2

41

10

2

R

I

!

µ"=

! and kB ˆ2

44

10

4

R

I

!

µ"=

!

Substitute for

2B

!and

4B

!to obtain:

ikjF ˆ

2

ˆ2

4

ˆ

2

2120

1

10

22sideleft

R

II

R

II

!µ

!µ !

!"

=""#

$%%&

'()(=

and

ikjF ˆ2

ˆ2

4ˆ

4

2140

4

1042sideright

R

II

R

II

!µ

!µ !

!"

"=##$

%&&'

(")=

Substitute numerical values and evaluate

sideleft F

!and sideright F

!:

( )( )( )( )

( )( )iiF ˆN100.1ˆ

m0200.02

A00.5A0.20m100.0N/A104 4

27

sideleft

!

!

"="

=#

#!

and ( )( )( )( )

( )( )iiF ˆN1029.0ˆ

m0700.02

A00.5A0.20m100.0N/A104 427

sideright

!

!

"!="

!=#

#!

(b) Express the net force acting on the coil:

sideright bottomsideleft topnet FFFFF

!!!!!+++=

Substitute for topF

!,

sideleft F

!,

bottomF

!, and sideright F

! and simplify to obtain:

( ) ( ) ( ) ( )

( )i

ijijF

ˆN1071.0

ˆN1029.0N 105.2ˆN100.1ˆN 105.2

4

4545

net

!

!!!!

"=

"!+"+"+"!=!