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Ch 8 Magnetic Field and Magnetic Forces

PC1432Peter Ho

Department of Physics, NUS

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8.1 Magnetic field due to a moving charge

A charge q moving in space has a magnetic field. If it has a velocity (that is much slower than the speed of light c), this magnetic field at any point in space (called the field point) in vacuum is found experimentally to be given by

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ˆ4 r

rxvqB o ⋅

πµ

=

v

is the unit vector (dimensionless) in the direction of . is the position vector of the field point from the electric charge.r

=πµ4

o 1x10–7 T m A–1 (by definition)

where µo is the permeability of free space, equal to 4π x 10–7 T m A–1

by definition.

r̂ r

r2 is the square of the distance between the charge and the location of interest.

[Later, you will see that 1 T m A–1 can also be written as 1 H m–1]

where,

is the velocity of the charge.v

v

+q

r

field point: where you wish todetermine the magnetic field

[Recall: The electric field of a point charge .]

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A uniformly moving electric charge is thus surrounded by a magnetic field, in addition to its electric field.

This magnetic field has the following properties:

(i) The magnetic field lines are circles centred on the line of motion and lie on planes perpendicular to the line of motion.

(ii) These lines have a sense given by the right-hand screw rule. Let your right-hand thumb point in the direction of charge motion corresponding to the conventional current, your curled fingers will give the sense of the encircling magnetic field lines.

(iii) The B magnitude decays as 1/r2 away from the charge (just as its electric field).

(iv) The B magnitude is proportional to the product of the speed of the charge and the magnitude of the charge.

v

+q

B

If the charge is accelerating, or moving near to the speed of light, the field lines get distorted. We won't study those cases here.

Side view

End view

X

B

Notation:= going into plane= coming out of the plane

X●

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Example. Forces between two moving charges in vacuum

(i) Two protons are moving on separate paths separated by distance d, parallel to each other, and in the same direction, and at the same speed v (<< c). Deduce the expression for the interaction force between them.Step 1. Identify the physics involved.Step 2. Set up the cartesian axes.Step 3. Sketch a diagram, and mark the fields and forces.Step 4. Perform the calculation.

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(ii) If these two protons are moving in opposite directions to each other but in still on parallel paths, deduce the expression for the interaction force at the instant when they pass each other.

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• If the magnetic field to be evaluated is not in vacuum but in some material medium, you only need to replace µo by µoµr, where µr is the relative permeability of the material. This µr takes into account the magnetisation of the material which may either reduce the magnetic field (if µr < 1) or enhance the field (if µr > 1).

Material µrCopper 1–1.0x10–5

Silver 1–2.6x10–5

Aluminium 1+2.2x10–5

Platinum 1+26x10–5

Nickel 100Steel 700Iron 4,000Permalloy (Ni-Fe) 8,000Mu-metal (Ni-Fe-Cu-Mo) 20,000

Thus except for ferromagnetic materials (which have µr >> 1), the rest of the materials have µrpractically equal to 1.000, and so µr can be neglected.

diamagnetic materials

paramagnetic materials

ferromagnetic materials

• Selected µr values of some materials:

= 0.99999

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8.2 Magnetic field due to a current element

The magnetic field in vacuum due to a moving charge element dq is 2

ˆ4 r

rxvdqBd o ⋅

πµ

=

As before, let us write

dIvdq ⋅=⋅

• We get the Biot–Savart law

2

ˆ4 r

rxdIBd o

⋅πµ

=

dI

r

is the length of the current element, with direction given by the conventional current.I is the magnitude of the conventional current.

is the position vector of the field point from the current element.r

r̂ is the unit vector (dimensionless) in the direction of . r

r2 is the square of the distance between the current element and the location of interest.

where,

d

current element

field point

The gives the magnetic field (at the field point) that is contributed by the current element.

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Every current element gives its own B field contribution to the field point. According to the superposition principle, the net B field is simply the sum of all these contributions. Each B field contribution is a vector, and so the sum is a vector sum.

• Therefore the net magnetic field generated by a current segment is given by the line integration of the Biot-Savart law over that current segment.

∫∫⋅

πµ

==line

or

rxdIBdB 2ˆ

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integral to sum up all the magnetic field contributions to the field point

line integral to sum up the magnetic field contribution of all the elements in the current line segment

• In special cases with simplifying symmetries, this can be solved by pencil and paper. In general, you need a computer to solve this integration.

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8.3 What is the magnetic field of a current in a infinite straight conductor?

∫⋅

πµ

=line

o

rrxdIB 2

ˆ4

)ˆ(sin4 2 k

rdyIB

y

y

o −θ⋅⋅

πµ

= ∫∞=

−∞=

In the coordinate system used,

• Therefore integration of the vector cross-product simplifies into integration of a scalar quantity representing the magnitude of the vector whose direction is constant,

rxd ˆ

is in the direction of , perpendicular to the xy-plane and pointing away from you. This direction does not depend on the location of the current element.

k̂−

ydd

=

)ˆ()(4 2/322 k

yxxdyIB

y

y

o −+

⋅⋅πµ

= ∫∞=

−∞=

r

I

x0

y

d

2

ˆ4 r

rxdIBd o

⋅πµ

=

θ

rx

=θsin2/122 )( yxr +=

• Set up the coordinate system as shown. Then apply Biot–Savart law,

• Look for simplifying symmetry. Art the field point marked on the xy-plane (see diagram)

which becomes

field point

B field contribution by current element:

curre

nt ele

ment

Step 1. Identify the physics involved.Step 2. Set up the cartesian axes.Step 3. Sketch a diagram, and mark the fields and forces.Step 4. Look for simplifying symmetry.Step 5. Perform the computation/ integration.(a) identify the integration variable(b) reduce the other variables to the integration variable

Note:

10

)ˆ()(4 2/322 k

yxdyxIB

y

y

o −+π

⋅⋅µ= ∫

∞=

−∞=

22/3222

)( xyxdyy

y

=+∫

∞=

−∞=

• From tables, e.g., S.O.S. Math, http://www.sosmath.com/tables/tables.html

)ˆ(2

kxIB o −⋅π⋅µ

=

• Hence

rIB o

⋅π⋅µ

=2

where r is the perpendicular distance between the conductor and field point.

I

B

• Question: What happens if the wire is not infinitely long?

The constant terms can come out of the integration sign to give

• This result is for the field point on the xy-plane. At other field points, the same consideration applies. Hence the B field has cylindrical symmetry. The field lines are concentric on the conductor with a sense given by the right-hand screw rule, and magnitude given by

you can get

[Recall: The electric field of an infinitely long line charge .]

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Properties of this field:

(i) The magnetic field lines are circles centred on the conductor and lie on planes perpendicular to this conductor.

(ii) These lines have a sense given by the right-hand screw rule. Let your right-hand thumb point in the direction of the corresponding conventional current, your four naturally curled fingers will give the sense of the encircling magnetic field lines.

(iii) The magnetic field magnitude decays as 1/r away from the current (which is slower than 1/r2 )

(iv) The magnetic field magnitude is proportional to the current.

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Example. Magnetic field due to a single long wireA long, straight conductor carriers a current of 1.0 A. Compute the magnetic field at a distance of 1.0 m from the wire.

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8.4 Magnetic force between currents in parallel conductors

dIB o

⋅π⋅µ

=2

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1222 BxIF

=

• The magnetic field set up by current1 has magnitude given by

and direction given by the right-hand screw rule. This will be perpendicular to the current in wire 2.

• You can check: (i) The force that the current in wire 2 exerts on current in wire 1 is equal and opposite to the force that current in wire 1 exerts on that in wire 2.

• Magnetic force experienced by a current length in wire 2 in the presence of a magnetic field set up by a current in wire 1 is given by

2

I1I2

1

2d

1B

2F

• Therefore d

IIF o 2212 2

⋅⋅πµ

=

• (ii) The force is attractive if the two currents are in the same direction, and repulsive if they are in the opposite direction.

1B

force on current 2 magnetic field set up by current 1

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This result is used to define the ampere: 1 A is the current which if present in two parallel conductors of infinite length separated by 1 m in air will cause each conductor to experience a force of exactly 2x10–7 N per unit length of the conductor.

Example. Magnetic force between two parallel currentsCompute the force per unit length between two infinite straight conductors carrying 1.0 A of current in the opposite directions.

1711772

172

102)(1021021

)1(1022

−−−−−−−− ===⋅=πµ

= NmxAmNAxTAxmATmAx

dIF o

Solution:

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gives a vector that lies on the xy-plane. This can be resolved into the axial and perpendicular components. The perpendicular component will be cancelled by the B field contributed by the current element on the opposite side of the loop. Therefore by symmetry only the axial component survives. This component has magnitude given by

8.5 Magnetic field of a circular current loop (on-axis)

2

ˆ4 r

rxdIBd o

⋅πµ

=

ra

=φcos2/122 )( axr +=

φ⋅⋅

πµ

= cos4 2// r

dIdB o

φ⋅⋅

πµ

= ∫ cos4 2//

loop

o

rdIB

r

d

I

φ

φ

a

x

y

0

Bd

//Bd

∫ +⋅⋅

πµ

=loop

o

axdaIB 2/322// )(4

• Integrating over the entire current loop,

• Set up the coordinate system as shown. Then apply Biot–Savart law,

• Look for simplifying symmetry. At the field point marked on the loop axis (see diagram),

rxd ˆ

//dB

which simplifies to

current element

field point

Note:

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2/322

2

// )(2 axaIB o

+⋅µ

=• Thus

• The direction of this magnetic field is again given by the right-hand screw rule: wrap the four fingers along the loop in the direction of the conventional current flow, then the thumb points in the direction of the axial magnetic field.

• At the centre of the loop, x = 0,a

IB o

2//⋅µ

=

• If the current coil is made of N tightly-packed turns, superposition of the contributions of each turn gives at the centre of the coil,

aNIB o

2//⋅⋅µ

=

• Qn: Where is the north of this magnetic dipole?

∫+⋅

πµ

=loop

o daxaIB 2/322// )(4

aaxaIB o π

+⋅

πµ

= 2)(4 2/322//which integrates to

IB

IB

IB

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8.6 Ampere's law

encloloop

IdB ⋅µ=⋅∫

For a steady-state electric current distribution, the integral of evaluated along any closed-loop integration path is given by the algebraic sum of the currents enclosed by the loop Iencl. Let us call the integral the “magnetic circulation”. The sense of this magnetic circulation is related to the sense of Iencl by the right-hand screw rule. For vacuum, this law is written as,

dB ⋅

This is a dot product given by θ⋅⋅=⋅ cos||||

dBdB

I1I2

1IdB oloop

⋅µ=⋅∫

)( 21 IIdB oloop

−⋅µ=⋅∫

)( 12 IIdB oloop

−⋅µ=⋅∫

0=⋅∫loop

dB

Zero here does not mean that B = 0, only the integral is zero.

dB ⋅

Examples:

The enclosed current is given by the net current that crosses any surface bounded by the loop.

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8.6 Applications of Ampere's lawAmpere's law can be quite useful to compute the magnetic field of highly symmetrical current distributions.

Procedure:Step 1. Select the integration path so that the magnetic field has a simple form by symmetry.Step 2. Perform the integration to evaluate the magnetic field.

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Example. Magnetic field inside a long cylindrical conductorA cylindrical non-magnetic conductor of radius R carries a current I uniformly distributed over its cross-sectional area (NB: This can be true only for dc currents. For ac currents, there is a skin effect which results in current being concentrated at the surface). Compute the magnetic field as a function of distance r from the axis of the conductor.

X X XX X X X X

X X X X XX X X

I

r • Select integration path. By symmetry the magnetic field lines must be circular and concentric on the axis of the conductor. You also know this from the Biot–Savart calculations. Therefore we choose concentric circular integration paths here.B

• Apply Ampere's law inside the conductor (r < R),

encloloop

IdB ⋅µ=⋅∫

2

2

2RrIrB o ⋅⋅µ=⋅π⋅which gives

Inside the conductor, the magnetic field increases linearly towards the surface.22 R

rIB o ⋅π⋅µ

=• Thus

• Apply Ampere’s law outside the conductor (r > R),IrB o ⋅µ=⋅π⋅2

rIB o

⋅π⋅µ

=2

which gives

(inside the conductor with uniform current)

(outside the conductor, same as by integration of Biot-Savart law)

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Example. Magnetic field inside a long solenoid

encloloop

IdB ⋅µ=⋅∫

• Select integration path. We already know from the Biot-Savart law that the B field of a current loop in the centre of the loop lies in the axial direction and has its direction given by the right-hand screw rule. • By symmetry, for an infinitely long solenoid of n turns per unit length, the magnetic field everywhere inside the solenoid must also lie in the axial direction. Outside the infinite solenoid, the magnetic field is zero. • Therefore we choose a rectangular integration path with one side aligned to the axial direction as shown.

⋅⋅⋅µ=⋅ nIB o

nIB o ⋅⋅µ=• Hence (magnetic field inside a long solenoid)

• This field is uniform everywhere inside the solenoid, and depends only on current and the number of turns per unit length, and not on the number of turns itself.

X X X X X X X X X X X X X X X X

• • • • • • • • • • • • • • • •

B = 0

B

n turns per unit length

I

• Apply Ampere’s law,

enclosed current is current times number of enclosed turns

IB

IB

IB

IB

IB

IB

By symmetry, the B field inside the solenoid lies in the axial direction.

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Summary

What you need to be able to do:(a) Compute or derive the magnetic field due to a moving electrical charge or electrical current using the Biot–Savart law, and hence the magnetic force on other moving electrical charge or electrical current.(b) Compute or derive the magnetic field of a steady-state current distribution using Ampere's law.