# Question 3 calculate current enclosed by the amperian

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Question 3:Calculate current enclosed by the Amperian loop. Be careful about signs!Remember thatJis non-uniform so you will need to compute the integralIenc=JˆndA∫∫.
Question 4:Your answer to Question 3 should be zero whenr=a.Whenr=b, the totalcurrent isIin the wire.Find an expression forI.
Question 5:What isBds?(That is, evaluate the integral, the left hand side ofAmpere’s law)
Question 6:Now apply Ampere’s Law.Write down a vector expressionfor the magneticfield in the regiona<r<b. Be sure to draw any necessary unit vectors on your abovefigure showing your Amperian loop.
Bds=μ0IencB(2πr)=μ02πα3(r3a3)ThereforeB=Bˆϕ=μ0α3(r3a3)1rˆϕ;a<r<b.Question 7:Repeat the steps above to find a vector expression for the magnetic field inthe regionr<a.0.
Question 8:Repeat the steps above to find a vector expression for the magnetic field inthe regionr>b.
Solving 7-4
The magnetic field for the regionr>bis given byB=Bˆϕ=μ0α3(b3a3)1rˆϕQuestion 9:Make aplot of the magnitude of the magnetic field vs. distance from thecentral axis, i.e.Bvs.r.Clearly label any special values.
Solving 7-5
Problem 2: Current SlabsThe figure below shows two slabs of current. Both slabs of current are infinite in thexandz-directions, and have thicknessdin they-direction. The top slab of current islocated in the region0<y<dand has a constant current densityJout=Jˆz, directed outof the plane of the figure below. The bottom slab of current is located in the regiond<y<0and has a constant current densityˆinJ=Jz, directed into the plane of thefigure below. There is a discontinuity in the current density aty=0(a)What is the direction and magnitude of the magnetic field foryd>?Brieflyexplain your reasoning. Clearly indicate your choice of unit vectors.-.
Solving 7-6

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Term
Fall
Professor
SCIOLLA
Tags
Magnetic Field, amperian loop