Chapter 27 Sources of Magnetic Field
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1 Chapter 27 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 is introduced and plays a role analogous to Gauss s Law. 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 where µ o = 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 The Forces Between Two Moving Protons 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. 3
4 F E = 1 ɛ o q 2 r 2 ĵ B = µ o q(v î) ĵ r 2 = µ o 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 r 2 ˆk = µ o q 2 v 2 r 2 The magnetic interaction in this situation is also repulsive. The ratio of the force magnitudes is ĵ F B F E = ɛ o µ o v 2 = v2 c 2 4
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 dq v d sin φ r 2 = µ o 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) r 2 d B = µ o 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 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
7 B = µ oi a a x dy (x 2 + y 2 ) 3/2 Integrating this by trigonometric substitution (y = x tan θ), we obtain: B = µ oi 2πx The magnetic field B has an axial symmetry about the y-axis. Magnetic Field of a Single Wire B = µ oi 2πr (5) Figure 5: This figure shows the magnetic-field around a long, straight, current-carrying conductor. The field lines are circles, with directions determined by the right-hand rule. 7
8 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. 8
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. 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. 9
10 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. (6) 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 the definition of the differential magnetic field d B produced by a current element I d l: db = µ oi dl (x 2 + a 2 ) Next, resolve the components of the db vector along the x and y directions 10
11 db x = db cos θ = µ oi dl a (x 2 + a 2 ) (x 2 + a 2 ) 1/2 db x = db sin θ = µ oi 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 a dl (x 2 + a 2 ) = µ oi 3/2 a (x 2 + a 2 ) 3/2 dl B x = µ o Ia 2 2 (x 2 + a 2 ) 3/2 (Magnetic field on the axis) (7) 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. 11
12 Figure 10: This figure shows the magnetic field strength along the axis of a circular coild 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. 12
13 6 Ampere s Law Ampere s Law for a Long, Straight Conductor Ampere s Law: General Statement 7 Applications of Ampere s Law Field of a Long, Straight, Current-Carrying Conducting Wire Field of a Long Cylindrical Conductor Field of a Solenoid Field of a Toroidal Solenoid The Bohr Magneton Paramagnetism Diamagnetism Ferromagnetism 13
14 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.. 14
15 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. 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. 15
16 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. 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. 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 16
17 Figure 18: This figure shows the magnetic field lines produced by the current in a solenoid. Figure 19: This figure shows how to calculate the strength of the magnetic field in a solenoid using Ampere s Law. 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. 17
18 Figure 21: This figure shows a toroidal solenoid with only a few turns shown. In parb (b) the integration paths (black circles) are used to computer the magnetic field B set up by the current (shown as dots and crosses). 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). 18
19 Figure 23: 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. Figure 24: 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. 19
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