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Unformatted text preview: Scott Hughes 30 March 2004 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Lecture 13: Ampere’s law revisited; Biot-Savart law Vector potential 13.1 Return to the magnetic field 13.1.1 What we did before Let’s quickly recap the major facts we’ve gone over regarding the magnetic field. A charge q moving with velocity ~v in a magnetic field ~ B feels a force ~ F = q ~v c × ~ B . Magnetic fields arise from flowing currents. The integral of ~ B along a closed path C the is related to the current enclosed by the path by Ampere’s law: I C ~ B · d~s = 4 π c I encl . If we apply this law to a long wire we find ~ B ( r ) = 2 I cr ˆ φ . where r is the distance to the wire and ˆ φ is related to the direction in which the current flows by the right-hand rule: point your right-hand thumb in the direction in which the current flows and ˆ φ curls around in the direction of your fingers. By placing many long wires next to each other and invoking superposition, we found the magnetic field of a sheet of current: L x θ y In the limit of L → ∞ , but holding the current per unit length in the sheet K ≡ I/L constant, we found that ~ B = + 2 πK c ˆ x above the plane, and ~ B =- 2 πK c ˆ x below the plane. The key observation here is that the change in the magnitude of the magnetic field as we cross the current is given by | ∆ ~ B | = 4 πK c . This should be reminiscent of the change in electric field when we cross a sheet of charge, | ∆ ~ E | = 4 πσ . Taking the divergence of the wire’s ~ B-field, we find ~ ∇· ~ B = 0 . Although we have only shown this holds for a specific example, this rule holds for all magnetic fields that arise from currents. Physically, it tells us that — as far we can tell — there is no such thing as an isolated magnetic charge. The smallest “chunks” of magnetic field that we can find come in a dipole configuration, with “north” and “south” poles. We never find an isolated “north” pole, with radial field lines. 13.1.2 Ampere’s law revisited Using Stoke’s theorem, we can rewrite the line integral of the ~ B field in terms of its curl: I C ~ B · d~s = Z S ( ~ ∇× ~ B ) · d~a . S is any surface bounded by the curve C . This is a nice result, since the current enclosed by C is given by an integral through the same surface S : I encl = Z S ~ J · d~a . Putting these two results together, Ampere’s law becomes Z S ( ~ ∇× ~ B ) · d~a = 4 π c Z S ~ J · d~a Z S µ ~ ∇× ~ B- 4 π c ~ J ¶ · d~a = 0 ....
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This note was uploaded on 04/11/2008 for the course PHYSICS 8.022 taught by Professor Hughes during the Spring '08 term at MIT.
- Spring '08