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5/16/16Oregon State University PH 213, Class #221 Here again is the field strength due to current in a straight wire: B =  0 I/(2  r) On a 3-D set of.

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Presentation on theme: "5/16/16Oregon State University PH 213, Class #221 Here again is the field strength due to current in a straight wire: B =  0 I/(2  r) On a 3-D set of."— Presentation transcript:

1 5/16/16Oregon State University PH 213, Class #221 Here again is the field strength due to current in a straight wire: B =  0 I/(2  r) On a 3-D set of coordinate axes: If a wire along the y-axis carries a current in the positive y-direction, which of these identical charges q, each moving with the same speed v, will experience the greatest magnetic force magnitude? 1. q at (2,0,0), moving parallel to the x-axis. 2. q at (3,0,0), moving parallel to the x-axis. 3. q at (3,0,0), moving parallel to the y-axis. 4. q at (2,0,0), moving parallel to the z-axis. 5. q at (1,0,0), moving parallel to the z-axis.

2 5/16/16Oregon State University PH 213, Class #222 Examples: A long, straight wire carries current, I. What is the magnitude of its magnetic force on … (a)a single charge q 0 moving at speed v parallel to the wire, at a distance r from that wire? F mag = q 0 v[  0 I/(2  r)] (b)another straight wire, of length L, carrying a current I 0, parallel to the wire carrying current I? F mag = I 0 L[  0 I/(2  r)]

3 What is the direction of the forces on wire 1? 1.Up from wire 2 and up from wire 3 2.Down from wire 2 and down from wire 3 3.Up from wire 2 and down from wire 3 4.Down from wire 2 and up from wire 3 5/16/163Oregon State University PH 213, Class #22

4 F = ILxB, and currents produce magnetic fields – so a wire will feel a force from the magnetic field DUE TO the OTHER wire B wire = (μ o /2π)(I/d) F 12 = I 2 L 2 xB 1 (3rd law pair symmetry) 5/16/164Oregon State University PH 213, Class #22

5 What is the current direction in this loop? And which side of the loop is the “north” pole? A.Current counterclockwise, north pole on bottom B.Current clockwise; north pole on bottom C.Current counterclockwise, north pole on top D.Current clockwise; north pole on top 5/16/165Oregon State University PH 213, Class #22

6 5/16/16Oregon State University PH 213, Class #226 The Magnetic Field of a Current in Wire Loops For the geometry of a single circular current loop of radius R, the magnetic field strength at the center of the loop is: B =  0 I/(2R) For N turns of wire all wrapped together in a flat loop, the field strength at the center is simply N times as great: B = N  0 I/(2R) For N turns of wire all wrapped together in a solenoid (helix) of length L, the field strength within the solenoid is: B =  0 NI/L (or B =  0 nI, where n is the number of turns per unit length). In all these cases, you would use RHR #2 to determine the direction of the B-field.

7 Find the strength and direction of B at point P. 5/16/167Oregon State University PH 213, Class #22

8 5/16/16Oregon State University PH 213, Class #228 Discovering and Defining the Magnetic Field The first observations of magnetism were of “permanent” magnets —certain materials or systems whose natural internal charge motions can sometimes align to cause magnetic fields. Examples: ・ Ferromagnetic metals (“bar magnets”) ・ The earth’s molten core We define the north pole of any bar magnet (e.g. a compass needle) as that end which would naturally point north when laid on the earth’s surface. The south pole is the other end. We then define magnetic field lines as emerging from the north pole, and terminating at the south pole, of a permanent bar magnet. Opposite poles of magnets attract; like poles repel.

9 5/16/16Oregon State University PH 213, Class #229 Ferromagnetism Currents are movements of charge. They can occur even at the atomic level: Electrons move around the protons in the nuclei; and they each exhibit “spin,” as well. Ordinarily, the random orientations of so many electrons’ motions will act to cancel out the B-field produced by any one motion. In certain materials, however, the energies and geometries of localized collections of electrons (called magnetic domains) don’t entirely result in their spin B-fields canceling one another. These small regions can act like many tiny net current loops— all aligned, like a solenoid. Such materials are ferromagnetic materials; and this phenomenon produces bar magnets. In fact, we can actually create such bar magnets, by inducing such micro-alignment of the electrons, using another, external magnetic field. This causes the magnetic domains to align with one another and to grow by influencing their neighbors.

10 What is the current direction in the loop? (The red arrows are showing the forces on both the magnet and the loop) 1.Out of the page at the top of the loop, into the page at the bottom 2.Out of the page at the bottom of the loop, into the page at the top 5/16/1610Oregon State University PH 213, Class #22

11 5/16/16Oregon State University PH 213, Class #2211 Geo-Magnetism Currents (movements of charge) can occur on much larger scales, as well. The earth’s interior contains many random collections of local net charge in motion (some moving in the molten portions of the core, but nearly all moving as the earth rotates, too). The overall effect is that of a large current loop—one that slowly (randomly) wobbles around, of course. In human history, the north pole of this current loop has been somewhat near the earth’s South (geographic) Pole; the south pole of this current loop is somewhat near the earth’s North (geographic) Pole. Thus the north pole ends of all our compass needles point basically north (why we named them as “north”): They are attracted to the south pole of the earth’s internal current loop.

12 Ampere’s Law Analog for B-field of Gauss’ law Requires symmetry to simplify the integration: wire, straight coil of wires (solenoid), or donut-shaped coil of wires (torus) Make a closed ‘Amperean loop’ – see how much current passes through it. Use right-hand rule #2 for loop direction, so that net current is out of the page. 5/16/1612Oregon State University PH 213, Class #22

13 The value of the line integral (above) around the closed path in the figure is 1.92×10 −5 Tm. What is the magnitude and direction of I 3 ? 1.0.7 A out of the page 2.0.7 A into the page 3.7.3 A out of the page 4.7.3 A into the page 5/16/1613Oregon State University PH 213, Class #22

14 5/16/1614Oregon State University PH 213, Class #22


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