6 Magnetostatics

6.1 The magnetic field

Although the phenomenon of magnetism was known about as early as the 13th century BC, and used incompasses it was only in 1819 than Hans Oersted recognised that magnetism and electricity are related(current in a wire deflects a compass needle).

Nowadays we know that magnetic fields are set up by charges in motion, as in

1. electromagnets - current flowing in a loop or coil of wire, and

2. permanent magnets - atomic-level current loops (circulating or spinning electrons) can, in somematerials, add together to give a net magnetism.

In electrostatics we introduced the electric field E to describe the electric force FE acting on a chargeq

FE = qE (6.1)

and we saw that the electric field surrounds any electric charge q1

E =1

4πε0

q1

r2r (6.2)

Equation 6.2 applies both to a stationary and a moving charge but in the latter case the region of spacesurrounding the moving charge contains both an electric field and a magnetic field B. This magnetic fieldexerts a magnetic force FB on any other charged particle q moving with a velocity v. Experiment showsthat �� ��FB = qv ×B (6.3)

where the cross-product shows that FB is perpendicular to both v and B. Note that FB never has a com-ponent parallel to v, so it cannot change the speed |v| or the kinetic energy 1

2mv2 of the charge. However

it can change the direction of travel of the charges.

The SI unit for B is the tesla. One tesla = one newton per (coulomb-metre per second) (1 T = 1 NC−1 m−1 s = 1 N A−1 m−1)

We can represent B with field lines, where

• the tangent to a field line gives the direction of B at that point, and

• the spacing of the field lines represents the magnitude of B.

Figure 55 shows the field lines near a bar magnet. Note that they exit one end (called the north pole)and enter at the other end (the south pole).

When two magnets are brought together, like magnetic poles repel each other, while opposite poles attract.

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Earth has its own magnetic field due to circulating molten metal in its core. A compass needle turnswith its north pole toward the Arctic. Although we call it the Earth’s (geographic) ‘north’ pole, it is infact the Earth’s magnetic south pole.

Typical field strengths:

Earth’s field 5× 10−4 TNear a small bar magnet 10−2 TLaboratory superconducting magnet 10 T

6.2 Motion of a charged particle in a magnetic field

1. For a uniform field, Figure 56 shows a charge moving in a plane perpendicular to B. Since themagnetic force is always at right angle to the motion, the force cannot increase the energy of the

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particle, and its speed |v| will remain constant. The particle adopts a circular path with the magneticforce

FB = qv ×B = −qvBr (6.4)

providing the necessary centripetal force

F centripetal = ma =mv2

rr (6.5)

to keep it there. Equating 6.4 and 6.5,mv2

r= qvB (6.6)

so the radius of the circle is ��

� r =

mv

qB(6.7)

and the period T (circumference/speed) is

T =2πr

v=

2π

v

mv

qB=

2πm

qB(6.8)

The angular frequency ω (the cyclotron frequency) of the motion is

ω = 2πf = 2π1

T=

qB

m(6.9)

independent of the speed of the particle.

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2. If the particle moves with velocity v at some arbitrary angle to the uniform B, its path is a helix (seeFigure 57). W can resolve the velocity of the particle into components parallel and perpendicular tothe field

v = v‖ + v⊥ (6.10)

The component v⊥ leads to circular motion at the cyclotron frequency. The component v‖ leads touniform motion parallel to B. The combined motion is helical.

3. For a non-uniform field, the motion can be more complex.

Figure 58 shows the Earth’s field, which channels charged particles (mostly from the solar wind) tothe magnetic poles. Here, the particles sometimes collide with atoms in the atmosphere and causethem to emit light, the Aurora Borealis and Aurora Australis.

6.3 The Lorentz force

A charge q moving with velocity v in both an electric field E and a magnetic field B experiences both anelectric force qE and a magnetic force qv ×B. The total force is therefore�� ��F = qE + qv ×B (6.11)

and is called the Lorentz force.

Look at three applications involving ‘crossed’ fields (E perpendicular to B).

1. The velocity selector

See Figure 59. Only those particles for which the downward force qE is matched by the upwardforce qvB will pass through the selector. Their speed is

��

� v =

E

B(6.12)

which can be selected by changing E or B.

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2. The mass spectrometer

In the instrument shown in Figure 60 a second uniform field B0 is placed beyond the end of a velocityselector. From equation 6.7, the radius of the semi-circular path in Figure 60 is

r =mv

qB0

(6.13)

so thatm

q=

rB0

v=

rB0B

E(6.14)

A variation of this method was used by J.J. Thomson in 1897 to measure me

efor electrons.

3. The Hall effect

In 1879, Edwin Hall saw that when a current-carrying conductor is placed in a magnetic field B, anelectric field EH is generated that is perpendicular to both B and vd (the drift velocity of the chargecarriers and the direction of current flow), see Figure 61.

It arises because the magnetic force pushes the charges to the edges of the conductor, leading to a‘Hall voltage’, ∆VH and hence and electric field EH = ∆VH

dwhere d is the width of the conductor.

We measure the sign of the charge carriers from the sign of ∆VH and the drift speed from themagnitude of EH. In equilibrium the magnetic force on the charge carriers and the electric force dueto the Hall field must balance

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qvdB = qEH (6.15)

therefore

vd =EH

B(6.16)

Once the drift velocity has been determined and knowing the dimensions of the conductor, the Hallvoltage can be used to measure magnetic field strength.

6.4 Magnetic force on a current-carrying conductor

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Consider a straight segment of current-carrying wire (length L, cross-sectional area A, current I - seeFigure 62) in a field B. The magnetic force on each charge q moving with drift velocity vd is qvd×B. Thetotal force on the segment is:

FB = nqALvd ×B (6.17)

where n is the number of charges per unit volume and therefore nqAL is the total charge in the volumeAL. Now, in terms of these quantities, the current is

I = nqvdA (6.18)

If we define a vector L as having magnitude L and the direction of vd (ie the direction of the current)we can write

FB = IL×B (6.19)

Now consider an arbitrarily shaped segment of wire.

From equation 6.19 the force on a small segment of vector length ds is

dFB = Ids×B (6.20)

and the total force on the wire is ��

� FB = I

∫ b

a

(ds×B) (6.21)

where a and b are the endpoints of the wire.

Note that if B is uniform we can take it out of the integral

FB = I

(∫ b

a

ds

)×B (6.22)

Furthermore, if the wire is a closed loop, ∫ b

a

ds = 0 (6.23)

so FB = 0. Thus the net magnetic force on a closed current loop in a uniform field is zero.

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6.5 Torque on a rectangular current loop; magnetic dipole

Much of the world’s work is done by electric motors. At the heart of a simple motor (Figure 63) is arectangular current-carrying loop in a uniform magnetic field B. What is the net torque on the loop?

If the long side of the loop has length a, and the short side length b, then looking end-on at Figure 63we see

For the sides of length a, the current is perpendicular to the magnetic field so the force has magnitude

|F 1| = |F 2| = IaB (6.24)

We define a unit vector n normal to the plane of the loop and with direction defined by a right-hand screwrule from the direction of the current. If the angle θ is the angle between n and B, the torque due to the

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forces F 1 and F 2 is

τ = F1b

2sin θ + F2

b

2sin θ (6.25)

= IabB sin θ (6.26)

= IAB sin θ (6.27)

where A = ab is the area of the loop.

Note that the forces on the short sides are equal in magnitude and opposite in direction.

A convenient form of equation 6.27 is obtained by defining the vector area A = An when�� ��τ = IA×B (6.28)

Equation 6.28 allows us also to calculate the potential energy U of a current loop in an external magneticfield. The torque tends to decrease the angle θ, therefore the potential energy increases with θ. This fixesthe sign between the derivative of the potential and the torque:

−∂U

∂θ= −τ = −IAB sin θ . (6.29)

By integrating we obtainU = −IAB cos θ + constant . (6.30)

If we choose the constant so that the potential energy U is zero when θ is π/2 we have

U = −IAB cos θ = −IA ·B . (6.31)

In analogy with the electric dipole moment, we define the magnetic dipole moment m of a small coilof area A carrying a current I to be �� ��m = IA (6.32)

where A is related to the current sens by the right-hand screw rule. In terms of the magnetic dipolemoment, the potential energy can be rewritten as�� ��U = −m ·B (6.33)

which is the magnetic analogous of the relation U = −p · E for the electric dipole.

6.6 The Biot-Savart law

So far we looked at the effect of a field B on moving charges and currents. Now we look at how currentsgive rise to B fields.

Following Oersted’s 1819 discovery of a compass needle deflection by a current-carrying conductor, Biotand Savart made quantitative measurements. They derived an expression for the field dB at a point Pproduced by a current element Ids:

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��

� dB =

µ0I

4π

ds× r

r2(6.34)

where r is a unit vector defined in the diagram andµ0 = 4π × 10−7 T m A−1 is a constant, known as the permeability of free space.

Equation 6.34 is the Biot-Savart law.

The total field at P is found by summing contributions from all the current elements��

� B =

µ0I

4π

∫ds× r

r2(6.35)

6.7 Magnetic field due to a current in a long straight wire

Consider a length element ds a distance r from a point P.

The cross product

ds× r = |ds||r| sin θ z (6.36)

= dx sin θ z (6.37)

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which on substitution in equation 6.34 gives

dB =µoI

4π

dx sin θ

r2z (6.38)

To integrate this we need to relate the variables r, θ, x. By geometry,

sin θ =a

r∴ r =

a

sin θ(6.39)

and

tan θ =a

−x∴ x =

−a

tan θ(6.40)

(the minus sign is needed since ds is at negative x).

∴ dx =adθ

sin2 θ(6.41)

and

dB =µ0I

4π

adθ

sin2 θsin θ

sin2 θ

a2(6.42)

=µoI

4πasin θdθ (6.43)

If the end points of the wire are at angles θ1 and θ2

then the total field is of magnitude

B =µ0I

4πa

∫ θ2

θ1

sin θdθ (6.44)

=µ0I

4πa(cos θ1 − cos θ2) (6.45)

If the wire is very long so that θ1 = 0 and θ2 = π, then cos θ1 − cos θ2 = 2 and��

� Bv. long wire =

µ0I

2πa(6.46)

The magnetic field lines (giving the direction of B) are circles concentric with the wire which lie inplanes perpendicular to the wire. The ‘curled-straight’ right-hand rule (Figure 64) can help us remember.

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6.8 Magnetic field due to a circular current loop

See Figure 65. Let us calculate B at an axial point P a distance x from the centre of a loop of radiusR.

Note that every ds on the loop is perpendicular to r. Therefore

|ds× r| = |ds||r| sin π

2= ds (6.47)

Also, note that r2 = x2 + R2. So, in the Biot-Savart law equation, the magnitude

dB =µ0I

4π

|ds× r|r2

=µ0I

4π

ds

(x2 + R2)(6.48)

The vector dB has components dBx and dBy, but in summing over the loop the contributions to dBy

cancel out. Thus B = Bxx, and

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Bx =

∮dBx =

∮dB cos θ =

µ0I

4π

∮ds cos θ

(x2 + R2)(6.49)

But θ, x,R are constants for all elements on the loop, so we can take them outside the integral, giving

Bx =µ0I

4π

cos θ

(x2 + R2)

∮ds (6.50)

But

cos θ =R√

x2 + R2and

∮ds = 2πR (6.51)

therefore ��

� Bx =

µ0IR2

2(x2 + R2)32

(6.52)

Let us now examine two extreme cases: the field at the center of the loop, and the field at a largedistance from the loop.

1. Field at the center of the loop.

At the centre of the loop, x = 0, so ��

� B0 =

µ0I

2R. (6.53)

2. Field at a large distance from the loop

In the denominator of Eq. 6.52 we can neglect R as x >> R. We obtain in this way

Bx 'µ0IR2

2x3. (6.54)

This can be rewritten in terms of the magnetic dipole moment m = IAn = IπR2n of the loop as��

� B =

µ0

4π

2m

x3. (6.55)

6.9 Magnetic force between two parallel conductors

Two current carrying wires will exert magnetic forces on each other (see Figure 66). Wire 2 creates a fieldB2 at wire 1, which exerts a force (see equation 6.19)

F 1 = I1l ×B2 (6.56)

As l is perpendicular to B2, |F 1| = F1 = I1lB2.

If the wires are very long, then from equation 6.46, B2 = µ0I22πa

and��

� F1 =

µ0I1I2l

2πa(6.57)

The direction of the force depends on the currents. If I1 is parallel to I2 the wires attract each other,while if I1 and I2 are anti-parallel, they repel each other.

Note that the force between two parallel wires is used to define the ampere.

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6.10 Ampere’s law

As we have seen, in magnetostatics (steady currents) we can find the the net magnetic field B due to anydistribution of currents via the Biot-Savart law

dB =µ0I

4π

ds× r

r2(6.58)

If the distribution is complicated we may need to use a computer to work out the total field. However,we can write equation 6.58 in a form which, in cases of high degrees of symmetry, is easily applied. Thisform is called Ampere’s law: �

��

∮B · ds = µ0I (6.59)

where I is the total current passing through any surface bounded by the closed path around which theline integral

∮B · ds is calculated.

We will see how this is applied to a wire, a solenoid and a toroid.

6.11 Magnetic field due to a current in a long straight wire

In section 6.7 we used the Biot-Savart law to find the field due to a long straight wire. After a trickyintegration we found that if the wire has a radius R and carries a current I0, then the field at a distancer from the axis of the wire is of magnitude

B(r) =µ0I0

2πr(r ≥ R) . (6.60)

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We can reach this result much more easily from Ampere’s law, since by symmetry (Figure 67)∮B · ds = B

∮ds = B.2πr = µ0I0 (6.61)

∴ B =µ0I0

2πr(r ≥ R) (6.62)

We can also use Ampere’s law inside the wire - path 2 in Figure 67 - where the current, I, enclosed is

I

I0

=πr2

πR2(6.63)

∴ I =r2

R2I0 (6.64)

Then applying Ampere’s law ∮B · ds = B.2πr =

r2

R2µ0I0 (6.65)

��

� ∴ B(r) =

µ0I0

2πR2r (r < R) (6.66)

so that the field magnitude varies as

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6.12 Magnetic field due to a solenoid

A tightly wound helical coil of wire is called a solenoid (see Figure 68). The magnetic field inside a solenoidis nearly uniform (Figure 69). Outside a solenoid the field is weak as the field contributions cancel eachother out. For an ideal solenoid (Figure 70), B = 0 outside and B is uniform inside. Ampere’s law thengives, for the path shown in Figure 70,∮

B · ds =

∫ b

a

B · ds +

∫ c

b

B · ds +

∫ d

c

B · ds +

∫ a

d

B · ds (6.67)

Now, along bc and da, B is perpendicular to ds, so the integrals are zero. Along cd, B = 0 and alongab, B is parallel to ds, so overall we have∮

B · ds =

∫ b

a

B · ds = Bh (6.68)

Then applying Ampere’s law

Bh = µ0Ienclosed = µ0NI (6.69)

where N is the number of turns in the solenoid. Thus we have for B,

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�� ��B = µ0nI (6.70)

where n = Nh

is the number of turns per unit length.

6.13 Magnetic field due to a toroid

A toroid is like a solenoid bent into the shape of a hollow ‘doughnut’ (Figure 71). By symmetry, theB field lines are concentric circles inside the toroid. Consider one of these as an Amperian path of radiusr, then ∮

B · ds = B.2πr = µ0NI (6.71)

��

� ∴ B =

µ0NI

2πr(6.72)

Note that for an ideal toroid, B = 0 anywhere outside the toroid - including in the ‘hole’ of the‘doughnut’.

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6.14 The magnetic field due to an infinite conducting sheet

One more example of using Ampere’s law,∮B · ds = µ0I (6.73)

to calculate B. Consider a thin, infinitely large sheet in the yz plane (Figure 73), carrying a current ofdensity J s per unit length in the +y direction.

By symmetry we expect B to be directed along ±z - think of it as a lot of long wires side by side(see Figure).

Applying Ampere’s law to the rectangular path abcd we get∮B · ds =

∮ b

a

B · ds +

∮ c

b

B · ds +

∮ d

c

B · ds +

∮ a

d

B · ds (6.74)

But

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∮ b

a

B · ds =

∮ d

c

B · ds = 0 because B is perpendicular to ds (6.75)

and ∮ c

b

B · ds =

∮ a

d

B · ds = Bl (6.76)

∴∮ b

a

B · ds = 2Bl = µ0Ienclosed = µ0Jsl (6.77)

��

� ∴ B =

µ0Js

2(6.78)

which is independent of the distance from the sheet ie it is uniform. Note the similarity with the uniformelectric field due to an infinite sheet of charge, E = σ

2ε0.

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