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In this chapter our main focus will be on Ampere’s law, a general theorem that allows us to calculate the magnetic fields of simple current distributions in much the same way that Gauss’ law allowed us to calculate the electric field of simple charge distributions. As we use them, Gauss’ and Ampere’s laws are integral theorems. With Gauss’ law.
Using Ampere’s Law, where we use the fact that the field is anti-symmetrical above and below the current sheet, and that the legs of the path perpendicular provide nothing to the integral: So, at a distance a beneath the sheet the magnetic field is: μ nI. in the positive x-direction. (Note there is no dependence on a.)
These notes give several examples of using the Ampere’s law to find the magnetic field. In the first example we reproduce the well-known magnetic field of an infinite straight wire from the Ampere’s law. We begin by using the symmetries of the wire — which we take to run along the z axis — to find the direction of the magnetic field.
two pieces of the same wire side by side? A battery of emf 6 volt is connected across 0.30 m of resistance wire of diameter 1.0 mm. A current of 0.5 ampere flows through the wire. What current will flow if this wire is replaced by 1.20 m of the same wire? A circuit consists of a 12 volt battery and two 4 ohm resistors connected in series.
Find the field contribution from a small element of current and then integrate over the current distribution. The current is uniformly distributed through the cross section of the wire. Since the wire has a high degree of symmetry, the problem can be categorized as a Ampère’s Law problem.
Solutions to Ampere’s Law Say we know the current distribution J(r) occurring in some physical problem, and we wish to find the resulting magnetic flux density B()r . Q: How do we find B(r) given J(r)? A: Two ways! We either directly solve the differential equation: ∇xr rBJ( )=µ 0 ( ) Or we first solve this differential equation for vector ...
Mass of an element produced at an electrode is proportional to the amount of electrical charge Q passed through the liquid. If a current of I Amperes (A) is passed through an electrolyte solution for t seconds (s), we have. where the units of Q is the Coulomb (C). Obviously, C = A × s.
