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Electricity and Magnetism Current Loops and Magnetic Dipoles Magnetism in Matter

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Electricity and Magnetism Current Loops and Magnetic Dipoles Magnetism in Matter Electricity and Magnetism Current Loops and Magnetic Dipoles Magnetism in Matter Lana Sheridan De Anza College Mar 5, 2018 Last time • magnetic field inside a solenoid • forces between current-carrying wires Overview • magnetic field around a current loop • more about magnetic dipoles • magnetism of matter Current loops and Magnetic dipoles We are now going to return to current loops and see why they have associated magnetic dipole moments, and how these behave. 908 Chapter 30 Sources of the Magnetic Field ▸ 30.2 continued Answer Yes, it can. The straight wires in Figure 30.4 do not contribute to the magnetic field. The only contribution is from the curved segment. As the angle u increases, the curved segment becomes a full circle when u 5 2p. Therefore, you can find the magnetic field at the center of a wire loop by letting u 5 2p in Equation 30.6: m I m I B 5 0 2p5 0 4pa 2a This result is a limiting case of a more general result discussed in Example 30.3. Magnetic field from a circular loop of wire In the last lecture, we considered the magnetic moment of a loop of wire, now we look at the magnetic field along a line through the Example 30.3 Magnetic Field on the Axiscenter of ofa Circular the loop. Current Loop y Consider a circular wire loop of radius a located in the yz plane and carrying a steady current I as in Figure 30.5. Cal- culate the magnetic field at an axial point P a distance x from the center of the loop. d Ss ˆr SOL U TION u a dBൿ S dB Conceptualize Compare this problem to Example 23.8 for O r the electric field due to a ring of charge. Figure 30.5 shows z S the magnetic field contribution d B at P due to a single cur- rent element at the top of the ring. This field vector can be x u I P x resolved into components dBx parallel to the axis of the ring dBx and dB! perpendicular to the axis. Think about the mag- netic field contributions from a current element atEach the bot little- segmentFigure 30.5 of wire (Example ds with 30.3) current GeometryI forcontributes calculating the a field dB. magnetic field at a point P lying on the axis of a current loop. tom of the loop. Because of the symmetry of the situation, S By symmetry, the total field B is along this axis. the perpendicular components of the field due toBy elements symmetry, we can see the the components that are parallel to at the top and bottom of the ring cancel. This cancellationthe plane of the ring will cancel. occurs for all pairs of segments around the ring, so we can ignore the perpendicular component of the field and focus solely on the parallel components, which simply add. Categorize We are asked to find the magnetic field due to a simple current distribution, so this example is a typical problem for which the Biot–Savart law is appropriate. Analyze In this situation, every length element d Ss is perpendicular to the vector r^ at the location of the element. Therefore, for any element, d Ss 3 r^ 5 ds 1 sin 9085ds. Furthermore, all length elements around the loop are at the same distance r from P, where r 2 5 a2 1 x 2. 0 0 1 21 2 S S m0I d s 3 r^ m0I ds Use Equation 30.1 to find the magnitude of d B dB 5 5 4p r 2 4p a 2 1 x 2 due to the current in any length element d Ss : 0 0 m I ds 1 2 Find the x component of the field element: dB 5 0 cos u x 4p a 2 1 x 2 1 m0I 2 ds cos u Integrate over the entire loop: B 5 dB 5 x C x 4p C a 2 1 x 2 a From the geometry, evaluate cos u: cos u5 a 2 1 x 2 1/2 m I1 ds 2 a m I a Substitute this expression for cos u into the inte- B 5 0 5 0 ds x 4p C a 2 1 x 2 a 2 1 x 2 1/2 4p a 2 1 x 2 3/2 C gral and note that x and a are both constant: c d 1 2 2 1 2 m0I a m0Ia Integrate around the loop: B 5 2pa 5 (30.7) x 4p a 2 1 x 2 3/2 2 a 2 1 x 2 3/2 1 2 1 2 1 2 908 Chapter 30 Sources of the Magnetic Field ▸ 30.2 continued Answer Yes, it can. The straight wires in Figure 30.4 do not contribute to the magnetic field. The only contribution is from the curved segment. As the angle u increases, the curved segment becomes a full circle when u 5 2p. Therefore, you can find the magnetic field at the center of a wire loop by letting u 5 2p in Equation 30.6: m I m I B 5 0 2p5 0 4pa 2a This result is a limiting case of a more general result discussed in Example 30.3. Magnetic field from a circular loop of wire Example 30.3 Magnetic Field on the Axis of a Circular Current Loop y Consider a circular wire loop of radius a located in the yz plane and carrying a steady current I as in Figure 30.5. Cal- culate the magnetic field at an axial point P a distance x from the center of the loop. d Ss ˆr SOL U TION u a dBൿ S dB Conceptualize Compare this problem to Example 23.8 for O r the electric field due to a ring of charge. Figure 30.5 shows z S the magnetic field contribution d B at P due to a single cur- rent element at the top of the ring. This field vector can be x u I P resolved into components dB parallel to the axis of the ring µ I ds ×^r x x dB = 0 dBx and dB! perpendicular to the axis. Think about the mag- 4π r 2 netic field contributions from a current element at the bot- Figure 30.5 (Example 30.3) Geometry for calculating the This time r ?magneticds field at a point P lying on the axis of a current loop. tom of the loop. Because of the symmetry of the situation, S By symmetry, the total field B is along this axis. the perpendicular components of the field due to elements at the top and bottom of the ring cancel. This cancellation ds ×^r = ds (cos θi + sin θj) occurs for all pairs of segments around the ring, so we can ignore the perpendicular component of the field and focus solely on the parallel components, which simply add. Categorize We are asked to find the magnetic field due to a simple current distribution, so this example is a typical problem for which the Biot–Savart law is appropriate. Analyze In this situation, every length element d Ss is perpendicular to the vector r^ at the location of the element. Therefore, for any element, d Ss 3 r^ 5 ds 1 sin 9085ds. Furthermore, all length elements around the loop are at the same distance r from P, where r 2 5 a2 1 x 2. 0 0 1 21 2 S S m0I d s 3 r^ m0I ds Use Equation 30.1 to find the magnitude of d B dB 5 5 4p r 2 4p a 2 1 x 2 due to the current in any length element d Ss : 0 0 m I ds 1 2 Find the x component of the field element: dB 5 0 cos u x 4p a 2 1 x 2 1 m0I 2 ds cos u Integrate over the entire loop: B 5 dB 5 x C x 4p C a 2 1 x 2 a From the geometry, evaluate cos u: cos u5 a 2 1 x 2 1/2 m I1 ds 2 a m I a Substitute this expression for cos u into the inte- B 5 0 5 0 ds x 4p C a 2 1 x 2 a 2 1 x 2 1/2 4p a 2 1 x 2 3/2 C gral and note that x and a are both constant: c d 1 2 2 1 2 m0I a m0Ia Integrate around the loop: B 5 2pa 5 (30.7) x 4p a 2 1 x 2 3/2 2 a 2 1 x 2 3/2 1 2 1 2 1 2 µ Ia = 0 (2πa) i 4π (a2 + x2)3=2 µ Ia2 = 0 i 2(a2 + x2)3=2 Similar to the E-field of an electric dipole... Magnetic field from a circular loop of wire (y-comp. cancels) µ cos θ B = 0 I ds i 4π r 2 p I a 2 2 Notice cos θ = r , and r = a + x . They are independent of the integration variable! µ Ia B = 0 ds i 4π (a2 + x2)3=2 I Magnetic field from a circular loop of wire (y-comp. cancels) µ cos θ B = 0 I ds i 4π r 2 p I a 2 2 Notice cos θ = r , and r = a + x . They are independent of the integration variable! µ Ia B = 0 ds i 4π (a2 + x2)3=2 I µ Ia = 0 (2πa) i 4π (a2 + x2)3=2 µ Ia2 = 0 i 2(a2 + x2)3=2 Similar to the E-field of an electric dipole..
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