Chapter 28 Sources of Magnetic Field

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1 Chapter 28 Sources of Magnetic Field In this chapter we investigate the sources of magnetic field, in particular, the magnetic field produced by moving charges (i.e., currents), Ampere s Law is introduced and plays a role analogous to Gauss s Law, and At the end of the chapter, various forms of magnetism produced in materials at the atomic and particle level are discussed (i.e., paramagnetism, diamagnetism, and ferromagnetism). 1

2 1 Magnetic Field of a Moving Charge Let s investigate the magnetic field due to a single point charge q moving with a constant velocity v. As we did for electric fields, the location of the moving charge at a given instant will be called the source point, and the point P where we want to calculate the magnetic field will be called the field point. Figure 1: This figure shows the magnetic-field vectors due to a moving positive point charge q. At each point, B is perpendicular to the plane of r and v. Notice that B points in the direction of v r. 2

3 1.1 Moving Charge: Vector Magnetic Field B = µ o 4π where µ o = 4π 10 7 T m/a (exactly). q v ˆr r 2 (1) A point charge in motion also produces an electric field, with field lines pointing radially outward from a positive charge. The magnetic field lines are completely different. The magnetic field lines are circles centered on the line of v and lying in planes perpendicular to this line. When we study the propagation of electromagnetic waves we will see that their speed of propagation is related to the electric permittivity and the magnetic permeability by the following relation: c 2 = 1 ɛ o µ o Ex A µC charge is moving at a constant speed of m/s in the +x-direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magnetic-field vector it produces at the following points: (a) x = m, y = 0, z = 0; (b) x = 0, y = m, z = 0; (c) x = m, y = m, z = 0; (d) x = 0, y = 0, z = m? The Forces Between Two Moving Protons F E = 1 4πɛ o q 2 r 2 ĵ B = µ o 4π q(v î) ĵ r 2 = µ o 4π qv r 2 ˆk The velocity of the upper proton is v = v î, so, the magnetic force on it is: F B = q( v) B = q( vî) µ o qv 4π r 2 3 ˆk = µ o q 2 v 2 4π r 2 ĵ

4 Figure 2: This figure shows the electric and magnetic forces acting on a pair of protons moving in opposite directions at their point of closest approach. The magnetic interaction in this situation is also repulsive. The ratio of the force magnitudes is F B F E = ɛ o µ o v 2 = v2 c 2 4

2 Magnetic Field of a Current Element Figure 3: This figure shows the magnetic-field vectors due to a current element d l. The view along the current axis is also shown.

5 2 Magnetic Field of a Current Element Figure 3: This figure shows the magnetic-field vectors due to a current element d l. The view along the current axis is also shown. dq = nqa dl The moving charges in this segment are equivalent to a single charge dq traveling with a velocity equal to the drift velocity v d. db = µ o 4π dq v d sin φ r 2 = µ o 4π n q v d A dl sin φ r 2 5

6 Current Element: Vector Magnetic Field This can be written in a more compact vector notion: db = µ o I dl sin φ (2) 4π r 2 d B = µ o 4π I d l ˆr r 2 (3) The total magnetic field due to a current-carrying conductor is the sum of the individual d B contributions: B = µ o 4π I d l ˆr r 2 (Biot-Savart Law) (4) 3 Magnetic Field of a Straight Current-Carrying Conductor Figure 4: This figure shows the magnetic field produced by a straight current-carrying conductor of length 2a. Using the Bio-Savart Law, we can calculate the magnetic field produced by a straight current-carrying conductor. 6

B = µ oi 4π a a x dy (x 2 + y 2 ) 3/2 Integrating this by trigonometric substitution (y = x tan θ), we obtain: B = µ oi 4π 2a x x 2 + a 2 (due to a finite wire of length 2a) (5) If we extend this

7 B = µ oi 4π a a x dy (x 2 + y 2 ) 3/2 Integrating this by trigonometric substitution (y = x tan θ), we obtain: B = µ oi 4π 2a x x 2 + a 2 (due to a finite wire of length 2a) (5) If we extend this solution for an infinitely long wire by taking the limit that a, we find the following magnetic field a distance r perpendicular the wire is: B = µ oi 2πr (due to an infinitely long wire) The magnetic field B has an axial symmetry about the y-axis (see Fig. 5 below) Magnetic Field of a Single Wire B = µ oi 2πr (6) Figure 5: This figure shows the magnetic-field around a long, straight, current-carrying conductor. The field lines form concentric circles, with directions determined by the right-hand rule. 7

8 Ex Currents in dc transmission lines can be 100 A or higher. Some people are concrned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percentage of the earth s magnetic field, which is 0.50 G. Is this value cause for worry? 8

9 Magnetic Field of Two Wires What is the magnetic field resulting from two long, straight current-carrying wires? We just use the principel of superposition. That is, B total = B 1 + B 2. Figure 6: This figure shows the magnetic-field lines for two long, straight conductors carrying equal currents in opposite directions. The magnetic field is strongest in the region between the conductors. 9

4 Force Between Parallel Conducting Wires Carrying Current Figure 7: This figure shows how the magnetic field B caused by the current in the lower conductor exerts a force F on the upper conductor.

10 4 Force Between Parallel Conducting Wires Carrying Current Figure 7: This figure shows how the magnetic field B caused by the current in the lower conductor exerts a force F on the upper conductor. The force between two parallel wires carrying current in the same direction are attractive, and consistent with Newton s 3 rd law. The magnetic field B produced by the lower conductor is: B = µ oi 2πr The force this field exerts on a length L of the upper wire is F = I L B, where the vector L is in the direction of the current I. 10

11 F = I LB = µ oii L 2πr This can be written in terms of the force per-unit-length: F L = µ o I I 2πr Magnetic Forces and Defining the Ampere One ampere is that unvarying current that, if present in each of two parallel conducting wires of infinite length and one meter apart in empty space, causes each conductor to experience a force of exactly newtons per meter of length. (7) 5 Magnetic Field of a Circular Current Loop Figure 8: This figure shows the magnetic-field vector d B produced by a current element I d l with components resolved along the x and y axes. We use Eq. 3 to determine the magnitude of d B produced by a current element I d l and find the following: db = µ oi 4π dl (x 2 + a 2 ) 11

Next, the components of the d B vector are resolved along the x and y directions, and we find the following: db x = db cos θ = µ oi 4π db y = db sin θ = µ oi 4π dl a (x 2 + a 2 ) (x 2 + a 2 ) 1/2 dl

12 Next, the components of the d B vector are resolved along the x and y directions, and we find the following: db x = db cos θ = µ oi 4π db y = db sin θ = µ oi 4π dl a (x 2 + a 2 ) (x 2 + a 2 ) 1/2 dl x (x 2 + a 2 ) (x 2 + a 2 ) 1/2 Integrating db y gives a null result. Meanwhile, integrating the x-component of db gives: B x = db x = µo I 4π a dl (x 2 + a 2 ) = µ oi 3/2 4π a (x 2 + a 2 ) 3/2 dl The integration d l = 2πa gives the follow result: B x = µ o Ia 2 2 (x 2 + a 2 ) 3/2 (Magnetic field on the axis) (8) Magnetic Field on the Axis of a Coil Figure 9: This figure shows how to apply the right-hand rule for the direction of the magnetic field produced on the axis of a current-carrying coil. The above equation can be written in terms of the magnetic dipole moment µ, where µ = NIπa 2 where N is the number of turns in the coil, and πa 2 is the area of the coil. 12

13 B x = µ o µ 2π (x 2 + a 2 ) 3/2 Figure 10: This figure shows the magnetic field strength along the axis of a circular coil with N turns. When x is much larger than a, the field magnitude decreases approximately as 1/x 3. Figure 11: This figure shows the magnetic field lines produced by the current in a circular loop. The B field along the x-axis has the same direction as the magnetic moment of the current loop µ. 13

14 6 Ampère s Law Gauss s law for electric fields involves the flux of E through a closed surface, and it states that the flux is equal to the total charge enclosed within the surface, divided by ɛ o. Strictly speaking Gauss s law for magnetic fields does not result in a useful relationship between magnetic fields and current distributions because it states that the flux of B through any closed surface is always zero, whether or not there are currents within the surface. Ampère s law is not formulated in terms of the magnetic flux, but rather in terms of the line integral of B around a closed path: C B d l Ampere s Law for a Long, Straight Conductor In order to develop Ampère s law, we revisit the magnetic field caused by a long, straight conductor carrying a current I. We found in a previous section that the magnetic field can be written as: B = µ o I 2πr This results in circular concentric magnetic field lines centered on the conductor (see Fig. 7 above). If we take the line integral of B around a circle with radius r, we find the following: C B d l = B dl = B dl = µ oi 2πr (2πr) = µ o I 14

15 Figure 12: This figure shows three integration paths for the line integral of B in the vicinity of a long, straight conductor carrying current I out of the plane of the page. The wire is viewed end-on. 15

16 Figure 13: This figure shows a more general integration path for the line integral of B around a long, straight conductor carrying current I out of the plane of the page. The wire is viewed end-on. Ampere s Law: General Statement Suppose we have several long, straight conductros passing through the surface bounded by the integration path C. The total magnetic field B at any point on the path is the vector sum of the fields produced by the individual currents. Thus the line integral of the total B equals µ o times the algebraic sum of the currents. If the integration path does not enclose a particular wire, the line integral of the B field due to that wire is zero. Thus we can replace I in our equation with I encl, the algebraic sum of the currents enclosed by the integration path, with the sum evaluated by using the sign rule. C B d l = µ o I encl (9) Ex A closed curve encircles several conductors. The line integral B d l around this curve is T m. (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite direction, what would be the values of the line integral? Explain 16

Figure 14: This figure shows the application of Ampere s law for multiple currents. You must have a high degree of symmetry in order to apply Ampere s law.

17 Figure 14: This figure shows the application of Ampere s law for multiple currents. You must have a high degree of symmetry in order to apply Ampere s law. In this case, the currents are approximated to be infinitely long. Figure 15: This figure shows two long, straight conductors carrying equal currents in opposite directions. The conductors are seen end-on, and the integration path is counterclockwise. The line integral B d l gets zero contribution from the upper and lower segments, a positive contribution from the left segment, and a negative contribution from the right segment; the net integral is zero. 17

7 Applications of Ampère s Law Ampère s law is useful when we can exploit symmetry to evaluate the line integral of B. We illustrate this with the following examples.

18 7 Applications of Ampère s Law Ampère s law is useful when we can exploit symmetry to evaluate the line integral of B. We illustrate this with the following examples. Field of a Long, Straight, Current-Carrying Conducting Wire B d l = Field of a Long Cylindrical Conductor B dl = B(2πr) = µ o I Figure 16: This figure shows how the magnetic field can be calculated inside a conductor r < R. The current through the gray area is (r 2 /R 2 )I. To find the magnetic field outside the conductor r > R, we apply Ampere s law to the circle enclosing the entire conductor. Inside the long cylindrical conductor we find that the current enclosed is (r 2 /R 2 )I. Evaluating the line integral B d l, we find B = µ oi 2π r R 2 (inside the conductor, r < R) Outside the conductor, we calculate again B d l = µo I encl, and we find the result we calculated before: B = µ oi 2πr (outside the conductor, r > R) 18

Figure 17: This figure shows the strength of the magnetic field inside and outside the conducting wire carrying a current I and having radius r = R Field of a Solenoid A solenoid is produced by

19 Figure 17: This figure shows the strength of the magnetic field inside and outside the conducting wire carrying a current I and having radius r = R Field of a Solenoid A solenoid is produced by wrapping an insulated conducting wire many times (N turns) around a form (e.g., a cylinder or toroid) in order to confine the magnetic field to particular region of space. Figure 18: This figure shows the magnetic field lines produced by the current in a solenoid. 19

20 Figure 19: This figure shows how to calculate the strength of the magnetic field in a solenoid using Ampere s Law. B d l = B L C B d l = B L = µ o N I where n = N/L, the number of turns per unit length. B = µ o n I (solenoid) (10) Figure 20: This figure shows the magnitude of the magnetic field at points along the axis of a solenoid with length 4a, equal to four times its radius a. The field magnitude at each end is about half its value at the center. 20

21 Ex A closely wound, circular coil with radius 2.40 cm has 800 turns. (a) What must the current in the coil be if the magnetic field at the center of the coil is T? (b) At what distance x from the center of the coil, on the axis of the coil is the magnetic field half its value at the center? 21

Field of a Toroidal Solenoid Figure 21: This figure shows a toroidal solenoid with only a few turns shown. In real applications there are many more turns.

22 Field of a Toroidal Solenoid Figure 21: This figure shows a toroidal solenoid with only a few turns shown. In real applications there are many more turns. In figure (b) there are three integration paths (black circles) used to compute the magnetic field B set up by the current (shown as dots and crosses). The figure above shows a doughnut-shaped toroidal solenoid, tightly wound with N turns of wire carrying a current I. Let s use Ampère s law to find the B-field inside and outside the toroid. Inside the toroid the line integral (path 2) encloses N times the current I. or C B d l = Nµ o I B 2πr = N µ o I B = µ o N I 2π r (toroidal solenoid) (11) 22

23 Ex A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80. (a) What is the magnetic field in the core? (b) What part of the magnetic field is due to atomic currents? 23

24 The Bohr Magneton In the Bohr model of the atom, electrons move in circular orbits about the nucleus. In some atoms, an external field can cause the current loops of valence electrons to become oriented preferentially with the field, so their magnetic fields add to the external field. We then say that the material is magnetized. Figure 22: This figure illustrates how an electron moving with speed v in a circular orbit of radius r has an angular moment L and an oppositely directed orbital magnetic dipole moment µ. The electron also has a spin angular momentum S and an oppositely directed spin magnetic dipole moment µ s (not shown in the figure). The electron s orbit produces both angular momentum L and a magnetic dipole moment µ. L = r p (angular momentum) µ = I A (magnetic dipole moment) Let s see how the magnetic dipole moment µ is related to the angular momentum L. As the electron orbits, it produces a current I = e/t = ev/2πr. µ = I A = ev 2πr πr2 = evr 2 = e 2m (mvr) = Strictly speaking this should be a vector equation, so, we write: e 2m L µ = e 2m L (12) 24

25 If the magnetic field defines the z direction, then µ z = e 2m L z In quantum mechanics, the angular momentum is observed to be quantized along the z-axis in units of h-bar where = h/2π = J s. where µ B = A m 2. µ B = e ( ) (Bohr magneton) (13) 2m Paramagnetism Diamagnetism Ferromagnetism 25

Figure 23: This figure shows the magnetic domains produced in a single crystal of nickel. The material becomes magnetized as the external magnetic field is increased.

26 Figure 23: This figure shows the magnetic domains produced in a single crystal of nickel. The material becomes magnetized as the external magnetic field is increased. The size of the domains change as the external magnetic field increasingly flips more of the dipole moments in each region to point in the direction of B, thus increasing the magnetization of the material. Figure 24: This figure shows the magnetization curve for a ferromagnetic material. The magnetization M approaches its saturation value M sat as the magnetic filed B o (caused by external currents) becomes large. 26

Chapter 28 Sources of Magnetic Field

Chapter 28 Sources of Magnetic Field In this chapter we investigate the sources of magnetic of magnetic field, in particular, the magnetic field produced by moving charges (i.e., currents). Ampere s Law