Chapter 30 - Magnetic Chapter 30 - Magnetic Fields and Torque Fields and Torque A PowerPoint Presentation by A PowerPoint Presentation by Paul E. Tippens, Professor Paul E. Tippens, Professor of Physics of Physics Southern Polytechnic State Southern Polytechnic State University University © 2007
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Chapter 30 - Magnetic Chapter 30 - Magnetic Fields and TorqueFields and Torque
A PowerPoint Presentation byA PowerPoint Presentation by
Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics
Southern Polytechnic State Southern Polytechnic State UniversityUniversity
A PowerPoint Presentation byA PowerPoint Presentation by
Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics
Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007
Objectives: Objectives: After completing After completing this module, you should be this module, you should be
able to:able to:• Determine the magnitude and Determine the magnitude and
direction of thedirection of the force force on a on a current-current-carrying wirecarrying wire in a in a BB-field-field..
• Calculate the Calculate the magnetic torquemagnetic torque on a on a coil or solenoid of area coil or solenoid of area AA, turns , turns NN, , and current and current I I in a given in a given B-fieldB-field. .
• Calculate the Calculate the magnetic fieldmagnetic field induced induced at the center of a at the center of a looploop or or coilcoil or at or at the interior of a the interior of a solenoidsolenoid..
Force on a Moving ChargeForce on a Moving ChargeRecall that the magnetic field Recall that the magnetic field BB in in teslas (T)teslas (T) was was defined in terms of the defined in terms of the force on a moving force on a moving chargecharge::
Recall that the magnetic field Recall that the magnetic field BB in in teslas (T)teslas (T) was was defined in terms of the defined in terms of the force on a moving force on a moving chargecharge::
sin
FB
qv sin
FB
qv
Magnetic Field
Intensity B:
1 N 1 N1 T
C(m/s) A m
1 N 1 N1 T
C(m/s) A m
B
vv
FF
SNN
B
vv
FF
BB
Force on a ConductorForce on a ConductorSince a current Since a current II is charge is charge qq moving through a moving through a wire, the magnetic force can be given in terms wire, the magnetic force can be given in terms
of current.of current.
Since a current Since a current II is charge is charge qq moving through a moving through a wire, the magnetic force can be given in terms wire, the magnetic force can be given in terms
of current.of current.
I = q/tL
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
FF Motion of +q
Right-hand rule: Right-hand rule: force F is force F is upward.upward.
F = qvBF = qvB
Since Since v = L/t, and I = q/t, v = L/t, and I = q/t, we can rearrange to find:we can rearrange to find:
L qF q B LB
t t
The The forceforce F F on a conductor of on a conductor of lengthlength L L and and currentcurrent II in in perpendicular perpendicular BB-field-field::
F = IBLF = IBL
Force Depends on Current Force Depends on Current AngleAngle
v sin I
B
v
F
Current I in wire: Length L
B
F = IBL sin
F = IBL sin
Just as for a moving Just as for a moving charge, the force on a charge, the force on a wire varies with wire varies with direction.direction.
Example 1.Example 1. A wire of length A wire of length 6 cm6 cm makes an angle makes an angle of of 202000 with a with a 3 mT3 mT magnetic field. What current is magnetic field. What current is needed to cause an upward force of needed to cause an upward force of 1.5 x 101.5 x 10-4-4
NN??
-4
-3 0
1.5 x 10 N
sin (3 x 10 T)(0.06m)sin20
FI
BL I = 2.44 AI = 2.44 A
Forces on a Current LoopForces on a Current LoopConsider a loop of area Consider a loop of area A = abA = ab carrying a carrying a current current I I in a constant in a constant BB field as shown field as shown below. below.
Consider a loop of area Consider a loop of area A = abA = ab carrying a carrying a current current I I in a constant in a constant BB field as shown field as shown below. below.
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
b
aI
The right-hand rule shows that the side forces The right-hand rule shows that the side forces cancel each other and the forces Fcancel each other and the forces F1 1 and Fand F22 cause a cause a
torque.torque.
The right-hand rule shows that the side forces The right-hand rule shows that the side forces cancel each other and the forces Fcancel each other and the forces F1 1 and Fand F22 cause a cause a
torque.torque.
n
A
B
SN
F2
F1
Normal vector
Torque
Torque on Current LoopTorque on Current Loop
x x x x x x x x x x x x x x x x x x x x
b
a I
2
a
2a
n
B
2 sina
2 sina
XF2
F1
Iout
Iin
Recall that Recall that torquetorque is product of is product of forceforce and and moment moment armarm..
The moment The moment arms for Farms for F1 1 and and FF2 2 are: are:
2 sina FF11 = F = F22 = IBb = IBb
21
22
( )( sin )
( )( sin )
a
a
IBb
IBb
22( )( sin ) ( )sinaIBb IB ab sinIBA sinIBA
In general, for a loop of In general, for a loop of NN turns carrying a current turns carrying a current II, we , we
have:have:sinNIBA sinNIBA
Example 2:Example 2: A A 200-turn200-turn coil of wire has a coil of wire has a radius of radius of 20 cm20 cm and the normal to the and the normal to the area makes an angle of area makes an angle of 303000 with a with a 3 mT3 mT B- B-field. What is the torque on the loop if the field. What is the torque on the loop if the current is current is 3 A3 A??
SN
nn
B
N = 200 turns
B = 3 mT; = 300
sinNIBA sinNIBA 2 2( .2m)A R
A = A = 0.126 m0.126 m22; N = 200 ; N = 200 turnsturns
B = 3 mT; B = 3 mT; = 30 = 3000; ; I = 3 I = 3 AA
2 0sin (200)(3 A)(0.003T)(0.126 m )sin 30NIBA
= 0.113 NmResultant torque on loop:Resultant torque on loop:
Magnetic Field of a Long Magnetic Field of a Long WireWire
When a current When a current II passes through a long straight passes through a long straight wire, the magnetic field wire, the magnetic field BB is is circularcircular as is shown as is shown by the pattern of iron filings below and has the by the pattern of iron filings below and has the indicated indicated directiondirection..
When a current When a current II passes through a long straight passes through a long straight wire, the magnetic field wire, the magnetic field BB is is circularcircular as is shown as is shown by the pattern of iron filings below and has the by the pattern of iron filings below and has the indicated indicated directiondirection..
Iron filingsI
B B
IThe right-hand The right-hand thumb rule:thumb rule: Grasp Grasp wire with right hand; wire with right hand; point thumb in point thumb in direction of direction of II. . Fingers Fingers wrap wire in direction wrap wire in direction of the of the circular circular B-field.B-field.
The right-hand The right-hand thumb rule:thumb rule: Grasp Grasp wire with right hand; wire with right hand; point thumb in point thumb in direction of direction of II. . Fingers Fingers wrap wire in direction wrap wire in direction of the of the circular circular B-field.B-field.
Calculating B-field for Long Calculating B-field for Long WireWire
The magnitude of the magnetic field The magnitude of the magnetic field B B at a at a distance distance rr from a wire is proportional to current from a wire is proportional to current II..
The magnitude of the magnetic field The magnitude of the magnetic field B B at a at a distance distance rr from a wire is proportional to current from a wire is proportional to current II..
0
2
IB
r
0
2
IB
r
Magnitude of Magnitude of BB-field -field for current for current II at at distance distance rr::The proportionality constant The proportionality constant is is
called the permeability of free called the permeability of free space:space:
Permeability: = 4x 10-7
Tm/A
B
I
rr
Circular B
X
Example 3:Example 3: A long straight wire carries A long straight wire carries a current of a current of 4 A4 A to the right of page. to the right of page. Find the magnitude and direction of the Find the magnitude and direction of the B-field at a distance of B-field at a distance of 5 cm5 cm above the above the wire.wire.
0
2
IB
r
0
2
IB
r
r = 0.05r = 0.05 m m II = 4 A = 4 A
-7 T mA(4 x 10 )(4 A)
2 (0.05 m)B
I = 4 Ar 5 cm
B=?
B = 1.60 x 10-5 T or 16 TB = 1.60 x 10-5 T or 16 T
I = 4 ArRight-hand Right-hand thumb rule:thumb rule: Fingers point Fingers point out out of paperof paper in in direction of B-direction of B-field.field.
Right-hand Right-hand thumb rule:thumb rule: Fingers point Fingers point out out of paperof paper in in direction of B-direction of B-field.field.
B out of
paper
Example 4:Example 4: Two parallel wires are Two parallel wires are separated by separated by 6 cm6 cm. Wire . Wire 11 carries a carries a current of current of 4 A4 A and wire and wire 22 carries a carries a current of current of 6 A6 A in the same direction. in the same direction. What is the resultant What is the resultant BB-field at the -field at the midpointmidpoint between the wires? between the wires?
0
2
IB
r
0
2
IB
r
I1 = 4 A
3 cmB=?
3 cm
I2 = 6 A
4 A
B1 out of
paper
1
6 A2
xB2 into paperBB11 is is
positivepositiveBB22 is is negativenegative
Resultant is Resultant is vector sum: vector sum: BBRR = =
BB
Example 4 (Cont.):Example 4 (Cont.): Find resultant B at Find resultant B at midpoint.midpoint.
I1 = 4 A
3 cmB=?
3 cm
I2 = 6 A
-7 T mA
1
(4 x 10 )(4 A)26.7 T
2 (0.03 m)B
-7 T mA
2
(4 x 10 )(6 A)40.0 T
2 (0.03 m)B
0
2
IB
r
0
2
IB
r
BB11 is is positivepositiveBB22 is is negativenegative
Resultant is vector sum: Resultant is vector sum: BBRR = =
BBBBRR = 26.7 = 26.7 T – 40 T – 40 T = -13.3 T = -13.3 TT
BBRR is into paper: is into paper: B = -13.3 T
Force Between Parallel Force Between Parallel WiresWires
I1
Recall wire with Recall wire with II11 creates creates BB11 at P:at P:
0 11 2
IB
d
0 11 2
IB
d
Out of paper!Out of paper!
d
P
I2d
Now suppose another wire with current Now suppose another wire with current II22 in same in same direction is parallel to first wire. Wire 2 experiences direction is parallel to first wire. Wire 2 experiences force force FF22 due to due to BB11. .
From right-hand rule, From right-hand rule, what is direction of what is direction of FF22??
Force F2
is Downward
Force F2
is Downward
F2
I2
F2
B
Parallel Wires (Cont.)Parallel Wires (Cont.)Now start with wire Now start with wire 2. 2. II22 creates creates BB22 at at
P:P:0 2
2 2
IB
d
0 22 2
IB
d
INTO INTO paper!paper!
Now the wire with current Now the wire with current II11 in same direction is in same direction is parallel to first wire. Wire 1 experiences force parallel to first wire. Wire 1 experiences force FF11 due due to to BB22. .
From right-hand rule, From right-hand rule, what is direction of what is direction of FF11??
Force F1
is Upward
Force F1
is Upward
I1
I1
F1 B
d
x
I22
B2 into paper
1d
PxF1
Parallel Wires (Cont.)Parallel Wires (Cont.)
I2d F1
I1
Attraction
I2d
F1
I1
Repulsion
F2
We have seen that We have seen that two parallel wires two parallel wires with currents in the with currents in the same direction are same direction are attracted to each attracted to each other.other.
We have seen that We have seen that two parallel wires two parallel wires with currents in the with currents in the same direction are same direction are attracted to each attracted to each other.other.
Use right-hand Use right-hand force rule to show force rule to show that oppositely that oppositely directed currents directed currents repel each other.repel each other.
Use right-hand Use right-hand force rule to show force rule to show that oppositely that oppositely directed currents directed currents repel each other.repel each other.
F2
Calculating Force on WiresCalculating Force on Wires
0 22 2
IB
d
0 22 2
IB
d
The field from The field from current in wire 2 is current in wire 2 is given by:given by:
The force The force FF11 on wire on wire 1 is:1 is:
F1 = I1B2LF1 = I1B2L
I2d F1
I1
Attraction
F2
1
2
L
0 21 1 2
IF I L
d
The same equation results The same equation results when considering Fwhen considering F22 due to due to BB11
The force per unit length The force per unit length for two wires separated for two wires separated by by d d is:is:
The force per unit length The force per unit length for two wires separated for two wires separated by by d d is:is:
0 1 2
2
I IF
L d
0 1 2
2
I IF
L d
Example 5:Example 5: Two wires Two wires 5 cm5 cm apart carry apart carry currents. The upper wire has currents. The upper wire has 4 A4 A north north and the lower wire has and the lower wire has 6 A6 A south. What south. What is the mutual force per unit length on is the mutual force per unit length on the wires?the wires?
I2 = 4 Ad=5 cmF
1
I1 = 6 A
Upper wire
F2
1
2
LLower wire
0 1 2
2
I IF
L d
0 1 2
2
I IF
L d
II11 = 6 A; = 6 A; II2 2 = 4 A; = 4 A; d d = 0.05 = 0.05 mmRight-hand rule applied to Right-hand rule applied to either wire shows either wire shows repulsionrepulsion..
-7 T mA(4 x 10 )(6 A)(4 A)
2 (0.05 m)
F
L
-59.60 x 10 N/mF
L
Magnetic Field in a Current Magnetic Field in a Current LoopLoop
NII IIB
Out
Right-hand rule shows Right-hand rule shows BB field directed out of field directed out of center.center.
Right-hand rule shows Right-hand rule shows BB field directed out of field directed out of center.center.
0
2
IB
R
Singl
e loop:
0
2
NIB
R
Coil of
N loops:
The SolenoidThe Solenoid
A A solenoidsolenoid consists of consists of many turns many turns NN of a wire of a wire in shape of a helix. The in shape of a helix. The magnetic magnetic B-fieldB-field is is similar to that of a bar similar to that of a bar magnet. The core can magnet. The core can be air or any material.be air or any material.
A A solenoidsolenoid consists of consists of many turns many turns NN of a wire of a wire in shape of a helix. The in shape of a helix. The magnetic magnetic B-fieldB-field is is similar to that of a bar similar to that of a bar magnet. The core can magnet. The core can be air or any material.be air or any material.
NS
Permeability
If the core is air: 4 x 10-7
Tm/A
If the core is air: 4 x 10-7
Tm/AThe The relative permeability relative permeability rr uses this value as a uses this value as a comparison.comparison.
00
or r r
00
or r r
The relative permeability for a medium ( r ):
The relative permeability for a medium ( r ):
The B-field for a SolenoidThe B-field for a Solenoid
For a solenoid of length For a solenoid of length LL, with , with NN turns and turns and current current II, the , the BB-field is -field is given by:given by:
For a solenoid of length For a solenoid of length LL, with , with NN turns and turns and current current II, the , the BB-field is -field is given by:given by:
NS
LSolenoid
NIB
L
NIB
L
Such a Such a BB-field -field is called the is called the magnetic inductionmagnetic induction since it arises or is produced by the current. It since it arises or is produced by the current. It applies to the interior of the solenoid, and its applies to the interior of the solenoid, and its directiondirection is given by the is given by the right-hand thumb ruleright-hand thumb rule applied to any current coil.applied to any current coil.
Such a Such a BB-field -field is called the is called the magnetic inductionmagnetic induction since it arises or is produced by the current. It since it arises or is produced by the current. It applies to the interior of the solenoid, and its applies to the interior of the solenoid, and its directiondirection is given by the is given by the right-hand thumb ruleright-hand thumb rule applied to any current coil.applied to any current coil.
Example 6:Example 6: A solenoid of length A solenoid of length 20 cm20 cm and 100 turns carries a current of and 100 turns carries a current of 4 A4 A. . The relative permeability of the core is The relative permeability of the core is 12,00012,000. What is the magnetic induction . What is the magnetic induction of the coil?of the coil?
N = 100 turns
20 cm
I = 4 A
7 T mA(12000)(4 x10 )
T mA0.0151
I = I = 4 A; 4 A; NN = 100 = 100 turns turns
0r L = 0.20 L = 0.20 m;m;
T mA(0.0151 )(100)(4A)
0.200 mB
A ferromagnetic core can significantly increase the B-field !
A ferromagnetic core can significantly increase the B-field !
B = 30.2 TB = 30.2 T
Summary of FormulasSummary of Formulas
I sin I
B
v
F
Current I in wire: Length L
B
F = IBL sin
F = IBL sin
The force F on a The force F on a wire carrying wire carrying current I in a given current I in a given B-field.B-field.
The force F on a The force F on a wire carrying wire carrying current I in a given current I in a given B-field.B-field.
sinNIBA sinNIBA
n
A
B
SN
F2
F1
The torque on a loop or The torque on a loop or coil of N turns and coil of N turns and current I in a B-field at current I in a B-field at known angle known angle ..
The torque on a loop or The torque on a loop or coil of N turns and coil of N turns and current I in a B-field at current I in a B-field at known angle known angle ..
Summary (Continued)Summary (Continued)
Permeability: Permeability: = 4= 4x 10x 10-7-7
TTm/Am/A
Permeability: Permeability: = 4= 4x 10x 10-7-7
TTm/Am/A
A circular magnetic field A circular magnetic field BB is is induced by a current in a wire. induced by a current in a wire. The direction is given by the The direction is given by the right-right-hand thumb rulehand thumb rule..
A circular magnetic field A circular magnetic field BB is is induced by a current in a wire. induced by a current in a wire. The direction is given by the The direction is given by the right-right-hand thumb rulehand thumb rule..
0
2
IB
r
The The magnitudemagnitude depends on the depends on the current current II and the and the distance distance rr from the from the wire.wire.
B
I
rr
Circular B
X
I
Summary (Continued)Summary (Continued)
The force per unit length The force per unit length for two wires separated for two wires separated by by d d is:is:
0 1 2
2
I IF
L d
0
2
IB
R
Singl
e loop:
0
2
NIB
R
Coil of
N loops:
For a solenoid of length For a solenoid of length LL, with , with NN turns and turns and current current II, the , the BB-field is -field is given by:given by:
NIB
L
CONCLUSION: Chapter 30CONCLUSION: Chapter 30Torque and Magnetic Torque and Magnetic
FieldsFields
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