NTNU Dpt. of Electronics Telecommunications, IME faculty, NTNU TFE420 Electromagnetics - rash course Lecture 4: Magnetic fields, Biot-avart s law, Ampere s law Thursday August th, 9-2 am. Problem. 0 min + 5 min solution. E = B t = 0, () E = ρ tot ɛ 0 (2) are the two laws concerning E in electrostatics. The corresponding laws for the magnetic field B are: E ɛ 0 µ 0 t + J tot, (3) 0. (4) In magnetostatics, i.e. if all currents are constant in time (so that the electromagnetic fields are time-independent) we have µ 0 J tot. (5) Biot-avart s law (the magnetic equivalent to oulomb s law) If we have to moving charges, Q moving with velocity v, Q 2 moving with v 2, the magnetic force on Q 2 from Q is F = Q 2 v 2 ( µ 0 Q v ˆr r 2 ), (6) where µ 0 = 0 7 Ns 2 2. This value is exact, because this equation is used as a stard for (the unit for charge). If v is parallel to v 2, v is normal to ˆr (the unit vector from Q to Q 2 ) we get F = µ 0 Q Q v 2v 2 ˆr. (7) r2 The magnetic force between two positive charges moving parallel to eachother is thus attractive.
We define the magnetic field B (the magnetic flux density) produced by a charge Q moving at v : µ 0 Q v ˆr r 2. (8) This is called Biot-avart s law. The magnetic field B describes the force on a charge Q moving with a velocity v: F = Qv B. (9) Example: harge moving in a constant B-field: The charge will follow a circular path. B-fields from different current distributions a) urrent along a line: d µ 0 Q dlv ˆr r 2, (0) is the contribution to the total B-field from a line segment dl along the thin conductor. We have Q dlv = dq dl dt = dq dl = Idl. () dt This gives d µ 0 Idl ˆr r 2. (2) This is the magnetic field B from a line current ( Biot-avart s law for a line current ). b) urrent along a surface: If the current instead is running along a thin surface we have This gives J = nq v = N qv = ρ s v. (3) d µ 0 J d ˆr r 2. (4) This is the magnetic field B from a surface current ( Biot-avart s law for a surface current ). c) urrent through a volume: which gives V J = Nqv, (5) d µ 0 V J ˆr r 3 dv. (6) This is the magnetic field B from a volume current ( Biot-avart s law for a volume current ). 2
Example: Magnetic field outside a cable with current I in the ŷ-direction. We find B at a given point (x, y) = (a, b): µ l/2 0 = µ 0 = µ 0 = µ 0aI l/2 l/2 = µ 0aI [ l/2 Idy ˆr r 2, sin θidy ˆφ r 2, aidy r 3 ˆφ, (a 2 + (b y) 2 dy ˆφ, ) 3/2 l 2 b a 2 a 2 + (b l 2 )2 + If we assume that the cable is long, i.e. l a, b we get l 2 + b a 2 a 2 + (b + l 2 )2 ]ˆφ. (7) µ 0I 2πa ˆφ. (8) Problem 2. 5 min + 5 min solution. Note: Using Biot-avart s law often involves complicated calculations. If the problem has some degree of symmetry we rather use Ampere s law. More about the B-field: The magnetic force does no work! dw = F mag dl = (qv B) dl dt = (qv B) vdt = 0. (9) dt Magnetic forces may change the direction, but not the velocity of a moving charge. There are no magnetic monopoles. It may be shown from Biot-avart s law that 0 (20) everywhere! This is one of Maxwell s equations. It means that magnetic fields cannot flow out from a point. That is, there are not magnetic charges. All magetic fields must bite their own tale. In integral form: BdV = B d = 0, (2) V i.e. the total magnetic flux through any closed surface is always zero! in fact we MUT do this, since we have ignored the return-current! 3
Amperes law for constant currents Applyling toke s theorem to (5) gives: B d = B dl = µ 0 J d. (22) This law tells us that the circulation of magnetic field around a closed path is proportional to the total current through the path. Example: Magnetic field outside a cable (again). B dl = µ 0 J d. (23) Let be a circular curve centered at the cable. We then have µ 0 J d = µ 0 Jd = µ 0 I, (24) B dl = B ˆφ dl ˆφ = Bdl = B dl = B(r)2πr. (25) We here used that B is directed along the ˆφ-direction, i.e. along the circular path around the cable. Further, we used that B is independent on φ z, thus only depends on r. ince r is constant along we could thus move B outside the integral. Equations (5), (24) (25) gives B(r) = µ 0I 2πr ˆφ, (26) which is the same as we got last time. We now argue why B(r, φ, z) = B(r)ˆφ. We assume that the cable is infinitely long. We thus don t have to bother about the returing current, or what happens at the end points. First, B must be independent on z, since the cable is infinitely long. onsidering B at different values for z does not alter the physical situation, thus our solution B(r, φ, z) should be independent on z. imilarly, (B) must be independent on φ, since if we rotate the cable by an angle φ wrt. the ẑ-axis, the physical situation is unaltered, so B(r, φ) must be independent on φ. If B has a ˆr-component, this will thus be equal for all φ z. This means we have a net flow of magnetic field out from a closed cylinder surface around the cable, which contradics (4). Finally, B cannot have a ẑ-component. This is seen from Biot-avart s law: µ 0 Idl ˆr r 2. (27) ince the current is along the ẑ-axis, we have Idl = Idlẑ, which gives no ẑ-component in B. Example 2: able with final thickness. Assume the current I runs through a cable with radius a, that the current is evenly distributed across the cross section: J = I ẑ. πa 2 Ampere s law gives B dl = µ 0 J d. (28) 4
As in the previous example, one may argue that B(r)ˆφ. This gives B dl = 2πrB(r). (29) The value of the right h side depends on r. For r > a we have J d = I, while for r < a we have Iπr2 J d =. This gives for r < a: πa 2 for r > a: µ 0Ir 2πa 2 ˆφ, (30) µ 0I 2πa ˆφ. (3) Problem 3 4. 45 min + 5 min solution. 5