Presentation on theme: "Magnetic Fields Due to Currents"— Presentation transcript:
1Magnetic Fields Due to Currents Magnetism Part IIMagnetic Fields Due to Currents
2Ampere’s LawThe line integral of the magnetic field B around any closed path is equal to μo times the net current across the area bound by the path.∫ B dl = μoI (integral around a closed loop)
3Using Ampere’s LawSelect the integration path. If you want to determine B at a point, then the path must pass through the point.The integration path doesn’t have to be any actual physical boundary. Usually it is a purely geometric curve.The integration path has to have enough symmetry to make evaluation of the integral possible.
4ContinuedThe B is tangent to all or some portion of the integration path and has the same magnitude of B at every point, then its integral equals Bl for that portion of the path.If B is perpendicular to all or some portion of the path, that portion makes no contribution to the integral.
5ContinuedThe sign of a current enclosed by the integration path is given by the right hand rule. Curl your fingers of your right hand so that they follow the integration path in the direction you carry out the integration. Your right thumb then points in the direction of positive current. If B is as described in step 4 and I enclosed is positive, then the direction of B is the same as the direction of the integration path: if instead I enclosed is negative, B is in the direction opposite that of integration.
6ContinuedIn the integral, B is always the total magnetic field at each point on the path. This field can be caused partly by currents enclosed by the path and partly by currents outside. If the path encloses no net current, the field points on the path need not be zero, but the integral is always zero.
7Long Straight WireA long straight wire of radius R carries a steady current I that is uniformly distributed through the cross section of the wire. Calculate the magnetic field a distance r from the center of the wire in the region r ≥ R and r < R
8Solutionr ≥ R∫ B dl = B ∫dl = B (2πr) = μoIB = μoI/(2πr)
9Solution r < R I’<I I’/I = π r2/πR2 I’ = (r2/R2) I ∫ B dl = B(2 πr) = μoI’B(2 πr) = μo(r2/R2)IB = μorI/(2πR2)
13ProblemA tightly wound solenoid has a length of 30 cm, a diameter of 2.0 cm, and contains 10,000 turns. If it carries a current of 5A, what is the magnitude of the magnetic field inside the solenoid?