Chapter 30 Chapter 30 -- Magnetic Fields Magnetic Fields and Torqueand Torque

A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

Objectives: Objectives: After completing this After completing this module, you should be able to:module, you should be able to:

•• Determine the magnitude and direction of Determine the magnitude and direction of thethe forceforce on a on a currentcurrent--carrying wirecarrying wire in a in a BB--fieldfield..

•• Calculate the Calculate the magnetic torquemagnetic torque on a coil or on a coil or solenoid of area solenoid of area AA, turns , turns NN, and current , and current I I in a given in a given BB--fieldfield. .

•• Calculate the Calculate the magnetic fieldmagnetic field induced at the induced at the center of a center of a looploop or or coilcoil or at the interior of or at the interior of a a solenoidsolenoid..

Force on a Moving ChargeForce on a Moving ChargeRecall that the magnetic field B in teslas (T) was defined in terms of the force on a moving charge: 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 chargeforce on a moving charge::

sinFB

qv

Magnetic Field Intensity B:

1 N 1 N1 TC(m/s) A m

Bvv

FF

SNN

Bvv

FF

BB

Force on a ConductorForce on a ConductorSince a current I is charge q moving through a wire, the magnetic force can be given in terms of current. Since a current Since a current II is charge is charge qq moving through a wire, moving through a wire, the magnetic force can be given in terms of current.the magnetic force can be given in terms 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

RightRight--hand rule: hand rule: force F is upward.force F is 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 LBt t

The The forceforce F F on a conductor of on a conductor of lengthlength LL and and currentcurrent II in in perpendicular perpendicular BB--fieldfield:: F = IBLF = IBL

Force Depends on Current AngleForce Depends on Current Angle

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 direction.wire varies with direction.

Example 1.Example 1. A wire of length A wire of length 6 cm6 cm makes an angle of makes an angle of 202000

with a with a 3 3 mTmT magnetic field. What current is needed to magnetic field. What current is needed to cause an upward force of cause an upward force of 1.5 x 101.5 x 10--44 NN??

-4

-3 0

1.5 x 10 Nsin (3 x 10 T)(0.06m)sin20FI

BL I = 2.44 AI = 2.44 A

Forces on a Current LoopForces on a Current LoopConsider a loop of area A = ab carrying a current I in a constant B field as shown below. Consider a loop of area Consider a loop of area A = A = abab carrying a current carrying a current I I in a constant in a constant BB field as shown below. field as shown 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

a I

The right-hand rule shows that the side forces cancel each other and the forces F1 and F2 cause a torque.

The rightThe right--hand rule shows that the side forces cancel hand rule shows that the side forces cancel each other and the forces Feach other and the forces F1 1 and Fand F22 cause a torque.cause a torque.

nA

B

SN

F2

F1Normal 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 armmoment arm..

The moment arms The moment arms for Ffor F1 1 and Fand F2 2 are:are:

2 sina FF11 = F= F22 = = IBbIBb

21

22

( )( sin )( )( sin )

a

a

IBbIBb

22( )( sin ) ( ) sinaIBb IB ab sinIBA

In general, for a loop of In general, for a loop of NN turns turns carrying a current carrying a current II, we have:, we have: sinNIBA

Example 2:Example 2: A A 200200--turnturn coil of wire has a radius coil of wire has a radius of of 20 cm20 cm and the normal to the area makes an and the normal to the area makes an angle of angle of 303000 with a with a 3 3 mTmT BB--field. What is the field. What is the torque on the loop if the current is torque on the loop if the current is 3 A3 A??

SN

nn

B

N = 200 turns

B = 3 mT;

= 300

sinNIBA 2 2( .2m)A R

A = A = 0.126 m0.126 m22; N = 200 turns; N = 200 turns

B = 3 B = 3 mTmT; ;

= 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 WireMagnetic Field of a Long Wire

When a current I passes through a long straight wire, the magnetic field B is circular as is shown by the pattern of iron filings below and has the indicated direction.

When a current When a current II passes through a long straight wire, the passes through a long straight wire, the magnetic field magnetic field BB is is circularcircular as is shown by the pattern of as is shown by the pattern of iron filings below and has the indicated iron filings below and has the indicated directiondirection..

Iron filingsI

B B

IThe right-hand thumb rule: Grasp wire with right hand; point thumb in direction of I. Fingers wrap wire in direction of the circular B-field.

The rightThe right--hand thumb hand thumb rule:rule: Grasp wire with Grasp wire with right hand; point thumb right hand; point thumb in direction of in direction of II. . Fingers Fingers wrap wire in direction of wrap wire in direction of the the circular circular BB--field.field.

Calculating BCalculating B--field for Long Wirefield for Long Wire

The magnitude of the magnetic field B at a distance r from a wire is proportional to current I. The magnitude of the magnetic field The magnitude of the magnetic field B B at a distance at a distance rr from a wire is proportional to current from a wire is proportional to current II..

0

2IBr

Magnitude of Magnitude of BB--field for field for current current II at distance at distance rr::

The proportionality constant The proportionality constant

is is called the permeability of free space:called the permeability of free space:

Permeability:

= 4

x 10-7 Tm/A

B

I

rr

Circular B

X

Example 3:Example 3: A long straight wire carries a A long straight wire carries a current of current of 4 A4 A to the right of page. Find the to the right of page. Find the magnitude and direction of the Bmagnitude and direction of the B--field at a field at a distance of distance of 5 cm5 cm above the wire.above the wire.

0

2IBr

r = 0.05r = 0.05 m m II = 4 A= 4 A

-7 TmA(4 x 10 )(4 A)

2 (0.05 m)B

I = 4 Ar

5 cmB=?

B = 1.60 x 10-5 T or 16 TB = 1.60 x 10-5 T or 16 T

I = 4 ArRight-hand thumb rule: Fingers point out of paper in direction of B-field.

RightRight--hand thumb hand thumb rule:rule: Fingers point Fingers point out of paperout of paper in in direction of Bdirection of B--field.field.

B out of paper

Example 4:Example 4: Two parallel wires are separated Two parallel wires are separated by by 6 cm6 cm. Wire . Wire 11 carries a current of carries a current of 4 A4 A and and wire wire 22 carries a current of carries a current of 6 A6 A in the same in the same direction. What is the resultant direction. What is the resultant BB--field at the field at the midpointmidpoint between the wires?between the wires?

0

2IBr

I1 = 4 A

3 cmB=?

3 cm

I2 = 6 A

4 A

B1 out of paper

1

6 A2

xB2 into paperBB11 is positiveis positive

BB22 is negativeis negative

Resultant is vector Resultant is vector sum: sum: BBRR = = BB

Example 4 (Cont.):Example 4 (Cont.): Find resultant B at midpoint.Find resultant B at midpoint.

I1 = 4 A

3 cmB=?

3 cm

I2 = 6 A

-7 TmA

1(4 x 10 )(4 A) 26.7 T

2 (0.03 m)B

-7 TmA

2(4 x 10 )(6 A) 40.0 T

2 (0.03 m)B

0

2IBr

BB11 is positiveis positive

BB22 is negativeis negative

Resultant is vector sum: Resultant is vector sum: BBRR = = BB

BBRR = 26.7 = 26.7 TT –– 40 40 TT = = --13.3 13.3 TT

BBRR is into paper:is into paper: B = -13.3 T

Force Between Parallel WiresForce Between Parallel Wires

I1

Recall wire with Recall wire with II11 creates creates BB11 at P:at P:

0 11 2

IBd

Out of paper!Out of paper!

d

PI2

d

Now suppose another wire with current Now suppose another wire with current II22 in same direction in same direction is parallel to first wire. Wire 2 experiences force is parallel to first wire. Wire 2 experiences force FF22 due to due to BB11 . .

From rightFrom right--hand rule, 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 2. Now start with wire 2.

II22 creates creates BB22 at P:at P:

0 22 2

IBd

INTO paper!INTO paper!

Now the wire with current Now the wire with current II11 in same direction is parallel to in same direction is parallel to first wire. Wire 1 experiences force first wire. Wire 1 experiences force FF11 due to due to BB22 . .

From rightFrom right--hand rule, 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

I22B2 into paper

1d

PxF1

Parallel Wires (Cont.)Parallel Wires (Cont.)

I2d F1 I1

Attraction

I2d

F1

I1

Repulsion F2

We have seen that two parallel wires with currents in the same direction are attracted to each other.

We have seen that We have seen that two parallel wires with two parallel wires with currents in the same currents in the same direction are attracted direction are attracted to each other.to each other.

Use right-hand force rule to show that oppositely directed currents repel each other.

Use rightUse right--hand force hand force rule to show that rule to show that oppositely directed oppositely directed currents repel each currents repel each other.other.

F2

Calculating Force on WiresCalculating Force on Wires

0 22 2

IBd

The field from current in The field from current in wire 2 is given by:wire 2 is given by:

The force FThe force F11 on wire 1 is:on wire 1 is: F1 = I1 B2 LF1 = I1B2 L

I2d F1 I1

Attraction

F2

1

2

L

0 21 1 2

IF I Ld

The same equation results The same equation results when considering Fwhen considering F22 due to Bdue to B11

The force per unit length for two wires separated by d is: The force per unit length for The force per unit length for two wires separated by two wires separated by d d is:is:

0 1 2

2I 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 and north and the lower wire has the lower wire has 6 A6 A south. What is the south. What is the mutual force per unit length on the wires?mutual force per unit length on the wires?

I2 = 4 Ad=5 cm F1 I1 = 6 A

Upper wire

F2

1

2

LLower wire

0 1 2

2I IF

L d

II11 = 6 A; = 6 A; II2 2 = 4 A; = 4 A; d d = 0.05 m= 0.05 m

RightRight--hand rule applied to 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)FL

-59.60 x 10 N/mFL

Magnetic Field in a Current LoopMagnetic Field in a Current Loop

NII IIB

Out

Right-hand rule shows B field directed out of center. RightRight--hand rule shows hand rule shows BB field directed out of center.field directed out of center.

0

2IB

R

Single loop:

0

2NIBR

Coil of N

loops:

The SolenoidThe Solenoid

A solenoid consists of many turns N of a wire in shape of a helix. The magnetic B- field is similar to that of a bar magnet. The core can be air or any material.

A A solenoidsolenoid consists of many consists of many turns turns NN of a wire in shape of a wire in shape of a helix. The magnetic of a helix. The magnetic BB-- fieldfield is similar to that of a is similar to that of a bar magnet. The core can bar 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/AIf the core is air:

4

x 10-7 Tm/A

The The relative permeability relative permeability rr uses this value as a comparison.uses this value as a comparison.

00

or r r

The relative permeability for a medium ( r ): The relative permeability for a medium ( r ):

The BThe B--field for a Solenoidfield for a Solenoid

For a solenoid of length L, with N turns and current I, the B-field is given by:

For a solenoid of length For a solenoid of length LL, , with with NN turns and current turns and current II, , the the BB--field is given by:field is given by:

NS

LSolenoid

NIBL

Such a B-field is called the magnetic induction since it arises or is produced by the current. It applies to the interior of the solenoid, and its direction is given by the right-hand thumb rule applied to any current coil.

Such a Such a BB--field field is called the is called the magnetic inductionmagnetic induction since it since it arises or is produced by the current. It applies to the arises or is produced by the current. It applies to the interior of the solenoid, and its interior of the solenoid, and its directiondirection is given by is given by the the rightright--hand thumb rulehand 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 and 100 turns carries a current of 100 turns carries a current of 4 A4 A. The . The relative permeability of the core is relative permeability of the core is 12,00012,000. . What is the magnetic induction of the coil?What is the magnetic induction 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 turns = 100 turns

0r L = 0.20 m;L = 0.20 m;

T mA(0.0151 )(100)(4 A)

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 wire carrying current I in a given B-field.

The force F on a wire The force F on a wire carrying current I in carrying current I in a given Ba given B--field.field.

sinNIBA

nA

B

SN

F2

F1 The torque on a loop or coil of N turns and current I in a B-field at known angle .

The torque on a loop or coil The torque on a loop or coil of N turns and current I in of N turns and current I in a Ba B--field at known angle field at known angle ..

Summary (Continued)Summary (Continued)

Permeability:

= 4

x 10-7 Tm/APermeability: Permeability:

= 4= 4

x 10x 10--77 TTm/Am/A

A circular magnetic field B is induced by a current in a wire. The direction is given by the right-hand thumb rule.

A circular magnetic field A circular magnetic field BB is induced is induced by a current in a wire. The direction is by a current in a wire. The direction is given by the given by the rightright--hand thumb rulehand thumb rule..

0

2IBr

The The magnitudemagnitude depends depends on the current on the current II and the and the distance distance rr from the wire.from the wire.

B

I

rr

Circular B

X

I

Summary (Continued)Summary (Continued)

The force per unit length for The force per unit length for two wires separated by two wires separated by d d is:is:

0 1 2

2I IF

L d

0

2IB

R

Single loop:

0

2NIBR

Coil of N

loops:

For a solenoid of length For a solenoid of length LL, , with with NN turns and current turns and current II, , the the BB--field is given by:field is given by:

NIBL

CONCLUSION: Chapter 30CONCLUSION: Chapter 30 Torque and Magnetic FieldsTorque and Magnetic Fields