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Transcript
6 Magnetostatics 6.1 The magnetic field Although the phenomenon of magnetism was known about as early as the 13th century BC, and used in compasses 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 some materials, add together to give a net magnetism. In electrostatics we introduced the electric field E to describe the electric force F E acting on a charge q F E = qE (6.1) and we saw that the electric field surrounds any electric charge q1 1 q1 r̂ (6.2) 4π0 r2 Equation 6.2 applies both to a stationary and a moving charge but in the latter case the region of space surrounding the moving charge contains both an electric field and a magnetic field B. This magnetic field exerts a magnetic force F B on any other charged particle q moving with a velocity v. Experiment shows that E= F B = qv × B (6.3) where the cross-product shows that F B is perpendicular to both v and B. Note that F B never has a component parallel to v, so it cannot change the speed |v| or the kinetic energy 12 mv 2 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 N C−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. 61 Earth has its own magnetic field due to circulating molten metal in its core. A compass needle turns with its north pole toward the Arctic. Although we call it the Earth’s (geographic) ‘north’ pole, it is in fact the Earth’s magnetic south pole. Typical field strengths: Earth’s field Near a small bar magnet Laboratory superconducting magnet 6.2 5 × 10−4 T 10−2 T 10 T 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 the magnetic force is always at right angle to the motion, the force cannot increase the energy of the 62 particle, and its speed |v| will remain constant. The particle adopts a circular path with the magnetic force (6.4) F B = qv × B = −qvBr̂ providing the necessary centripetal force F centripetal = ma = mv 2 r̂ r (6.5) to keep it there. Equating 6.4 and 6.5, mv 2 = qvB r (6.6) so the radius of the circle is mv r= qB (6.7) and the period T (circumference/speed) is T = 2πr 2π mv 2πm = = v v qB qB (6.8) The angular frequency ω (the cyclotron frequency) of the motion is ω = 2πf = 2π independent of the speed of the particle. 63 1 qB = T m (6.9) 2. If the particle moves with velocity v at some arbitrary angle to the uniform B, its path is a helix (see Figure 57). W can resolve the velocity of the particle into components parallel and perpendicular to the field (6.10) v = vk + v⊥ The component v ⊥ leads to circular motion at the cyclotron frequency. The component v k leads to uniform 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) to the magnetic poles. Here, the particles sometimes collide with atoms in the atmosphere and cause them 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 an electric 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 upward force qvB will pass through the selector. Their speed is E v= B which can be selected by changing E or B. 64 (6.12) 2. The mass spectrometer In the instrument shown in Figure 60 a second uniform field B 0 is placed beyond the end of a velocity selector. From equation 6.7, the radius of the semi-circular path in Figure 60 is r= mv qB0 (6.13) so that m rB0 rB0 B = = q v E A variation of this method was used by J.J. Thomson in 1897 to measure (6.14) me e for electrons. 3. The Hall effect In 1879, Edwin Hall saw that when a current-carrying conductor is placed in a magnetic field B, an electric field E H is generated that is perpendicular to both B and v d (the drift velocity of the charge carriers 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 = ∆Vd H where 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 the magnitude of EH . In equilibrium the magnetic force on the charge carriers and the electric force due to the Hall field must balance 65 qvd B = qEH (6.15) therefore vd = EH B (6.16) Once the drift velocity has been determined and knowing the dimensions of the conductor, the Hall voltage can be used to measure magnetic field strength. 6.4 Magnetic force on a current-carrying conductor 66 Consider a straight segment of current-carrying wire (length L, cross-sectional area A, current I - see Figure 62) in a field B. The magnetic force on each charge q moving with drift velocity v d is qv d × B. The total force on the segment is: F B = nqALv d × B (6.17) where n is the number of charges per unit volume and therefore nqAL is the total charge in the volume AL. Now, in terms of these quantities, the current is I = nqvd A (6.18) If we define a vector L as having magnitude L and the direction of v d (ie the direction of the current) we can write F B = 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 dF B = Ids × B and the total force on the wire is Z b (ds × B) FB = I (6.20) (6.21) a where a and b are the endpoints of the wire. Note that if B is uniform we can take it out of the integral Z b FB = I ds × B (6.22) a Furthermore, if the wire is a closed loop, Z b ds = 0 (6.23) a so F B = 0. Thus the net magnetic force on a closed current loop in a uniform field is zero. 67 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 a rectangular 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 63 we 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 n̂ normal to the plane of the loop and with direction defined by a right-hand screw rule from the direction of the current. If the angle θ is the angle between n̂ and B, the torque due to the 68 forces F 1 and F 2 is b b τ = F1 sin θ + F2 sin θ 2 2 = IabB sin θ = IAB sin θ (6.25) (6.26) (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 = An̂ 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 magnetic field. The torque tends to decrease the angle θ, therefore the potential energy increases with θ. This fixes the sign between the derivative of the potential and the torque: − ∂U = −τ = −IAB sin θ . ∂θ (6.29) By integrating we obtain U = −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 coil of 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 dipole moment, 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 currents give rise to B fields. Following Oersted’s 1819 discovery of a compass needle deflection by a current-carrying conductor, Biot and Savart made quantitative measurements. They derived an expression for the field dB at a point P produced by a current element Ids: 69 µ0 I ds × r̂ (6.34) dB = 4π r2 where r̂ is a unit vector defined in the diagram and µ0 = 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 Z µ0 I ds × r̂ (6.35) B= 4π r2 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 × r̂ = |ds||r̂| sin θ ẑ = dx sin θ ẑ 70 (6.36) (6.37) which on substitution in equation 6.34 gives µo I dx sin θ ẑ 4π r2 To integrate this we need to relate the variables r, θ, x. By geometry, a a sin θ = ∴r= r sin θ and a −a tan θ = ∴x= −x tan θ (the minus sign is needed since ds is at negative x). dB = ∴ dx = adθ sin2 θ (6.38) (6.39) (6.40) (6.41) and µ0 I adθ sin2 θ sin θ 4π sin2 θ a2 µo I = sin θdθ 4πa If the end points of the wire are at angles θ1 and θ2 dB = (6.42) (6.43) then the total field is of magnitude Z µ0 I θ2 B = sin θdθ 4πa θ1 µ0 I = (cos θ1 − cos θ2 ) 4πa If the wire is very long so that θ1 = 0 and θ2 = π, then cos θ1 − cos θ2 = 2 and Bv. long wire µ0 I = 2πa (6.44) (6.45) (6.46) The magnetic field lines (giving the direction of B) are circles concentric with the wire which lie in planes perpendicular to the wire. The ‘curled-straight’ right-hand rule (Figure 64) can help us remember. 71 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 radius R. Note that every ds on the loop is perpendicular to r̂. Therefore |ds × r̂| = |ds||r̂| sin π = ds 2 (6.47) Also, note that r2 = x2 + R2 . So, in the Biot-Savart law equation, the magnitude dB = µ0 I |ds × r̂| µ0 I ds = 2 2 4π r 4π (x + 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 = Bx x̂, and 72 I I I µ0 I ds cos θ (6.49) Bx = dBx = dB cos θ = 4π (x2 + R2 ) But θ, x, R are constants for all elements on the loop, so we can take them outside the integral, giving I µ0 I cos θ Bx = ds (6.50) 4π (x2 + R2 ) But I R cos θ = √ and ds = 2πR (6.51) x2 + R 2 therefore µ0 IR2 Bx = (6.52) 3 2(x2 + R2 ) 2 Let us now examine two extreme cases: the field at the center of the loop, and the field at a large distance from the loop. 1. Field at the center of the loop. At the centre of the loop, x = 0, so µ0 I B0 = . 2R 2. Field at a large distance from the loop (6.53) In the denominator of Eq. 6.52 we can neglect R as x >> R. We obtain in this way µ0 IR2 Bx ' . (6.54) 2x3 This can be rewritten in terms of the magnetic dipole moment m = IAn̂ = IπR2 n̂ of the loop as µ0 2m B= . 4π x3 6.9 (6.55) Magnetic force between two parallel conductors Two current carrying wires will exert magnetic forces on each other (see Figure 66). Wire 2 creates a field B 2 at wire 1, which exerts a force (see equation 6.19) F 1 = I1 l × B 2 (6.56) As l is perpendicular to B 2 , |F 1 | = F1 = I1 lB2 . If the wires are very long, then from equation 6.46, B2 = µ0 I2 2πa and µ0 I1 I2 l F1 = (6.57) 2πa The direction of the force depends on the currents. If I 1 is parallel to I 2 the wires attract each other, while if I 1 and I 2 are anti-parallel, they repel each other. Note that the force between two parallel wires is used to define the ampere. 73 6.10 Ampère’s law As we have seen, in magnetostatics (steady currents) we can find the the net magnetic field B due to any distribution of currents via the Biot-Savart law dB = µ0 I ds × r̂ 4π 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. This form is called Ampère’s law: I B · ds = µ0 I (6.59) where I is the H total current passing through any surface bounded by the closed path around which the line 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 tricky integration we found that if the wire has a radius R and carries a current I0 , then the field at a distance r from the axis of the wire is of magnitude B(r) = µ0 I0 2πr 74 (r ≥ R) . (6.60) We can reach this result much more easily from Ampère’s law, since by symmetry (Figure 67) I I (6.61) B · ds = B ds = B.2πr = µ0 I0 ∴B = µ0 I0 (r ≥ R) 2πr (6.62) We can also use Ampère’s law inside the wire - path 2 in Figure 67 - where the current, I, enclosed is πr2 I = I0 πR2 ∴I= (6.63) r2 I0 R2 (6.64) Then applying Ampère’s law I B · ds = B.2πr = r2 µ0 I0 R2 (6.65) µ0 I0 ∴ B(r) = r 2πR2 (r < R) so that the field magnitude varies as 75 (6.66) 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 solenoid is nearly uniform (Figure 69). Outside a solenoid the field is weak as the field contributions cancel each other out. For an ideal solenoid (Figure 70), B = 0 outside and B is uniform inside. Ampère’s law then gives, for the path shown in Figure 70, I Z b Z c Z d Z a B · ds + B · ds + B · ds + B · ds (6.67) B · ds = a b c d Now, along bc and da, B is perpendicular to ds, so the integrals are zero. Along cd, B = 0 and along ab, B is parallel to ds, so overall we have Z b I B · ds = Bh (6.68) B · ds = a Then applying Ampère’s law Bh = µ0 Ienclosed = µ0 N I where N is the number of turns in the solenoid. Thus we have for B, 76 (6.69) B = µ0 nI where n = 6.13 N h (6.70) is the number of turns per unit length. Magnetic field due to a toroid A toroid is like a solenoid bent into the shape of a hollow ‘doughnut’ (Figure 71). By symmetry, the B field lines are concentric circles inside the toroid. Consider one of these as an Ampèrian path of radius r, then I B · ds = B.2πr = µ0 N I (6.71) µ0 N I ∴B= 2πr (6.72) Note that for an ideal toroid, B = 0 anywhere outside the toroid - including in the ‘hole’ of the ‘doughnut’. 77 6.14 The magnetic field due to an infinite conducting sheet One more example of using Ampère’s law, I B · ds = µ0 I (6.73) to calculate B. Consider a thin, infinitely large sheet in the yz plane (Figure 73), carrying a current of density 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 Ampère’s law to the rectangular path abcd we get I I b I c I d I B · ds = B · ds + B · ds + B · ds + a b c But 78 d a B · ds (6.74) I b I d B · ds = a B · ds = 0 because B is perpendicular to ds (6.75) c and I c I a B · ds = b I B · ds = Bl b B · ds = 2Bl = µ0 Ienclosed = µ0 Js l ∴ (6.76) d (6.77) a µ0 J s ∴B= 2 (6.78) which is independent of the distance from the sheet ie it is uniform. Note the similarity with the uniform electric field due to an infinite sheet of charge, E = 2σ0 . 79