Collaborative ASSIGNMENT Assignment 3: Sources of magnetic fields Solution

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Walter Watkins
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1 Electicity and Magnetism: PHY Novembe, 014 Collaboative ASSIGNMENT Assignment 3: Souces of magnetic fields Solution 1. a A conducto in the shape of a squae loop of edge length l m caies a cuent I as shown in the figue below. Calculate the magnitude and diection of the magnetic field at the cente of the squae. bwhat If? If this conducto is fomed into a single cicula tun and caies the same cuent, what is the value of the magnetic field at the cente? d 90+θ l/ θ Field inwads. x y Due Date: 18 Novembe, 014, 5:00 pm 1

2 Electicity and Magnetism: PHY Novembe, 014 db = µ 0 = I d l ˆ + l d sin π + θ + l µ 0 I + l cos θ d. Now l/ = tan θ, = l tan θ d = l sec θ dθ. db = l cos θ dθ πl B πl µ 0 I l sec θ cos θ sec θ dθ + tan 1 l/ θ= tan 1 l/ πl sin θ +π/4 π/4 = µ 0I πl cos θ dθ πl Fields ae additive in this case Futhemoe a wie of length 4l implies a cicle of adius l π. l π Field at its cente is, B = µ 0 I = µ 0πI 4l π ϕ=0 π l π l dϕ π l π Due Date: 18 Novembe, 014, 5:00 pm

Electicity and Magnetism: PHY-04. 11 Novembe, 014. A conducto consists of a cicula loop of adius R and two staight, long sections as shown in the figue below.

3 Electicity and Magnetism: PHY Novembe, 014. A conducto consists of a cicula loop of adius R and two staight, long sections as shown in the figue below. The wie lies in the plane of the pape and caies a cuent I. Find the vectoial expession fo the magnetic field at the cente of the loop. Fom infinite wie, B 1, into the plane of the pape. πr Fom cicle, B, also into the plane of the pape. R see the pevious question. Hence B πr + µ 0I R, into the plane of the pape. 3. A sphee of adius R has a unifom volume chage density ρ. Detemine the magnetic field at the cente of the sphee when it otates as a igid object with angula speed ω about an axis though its cente, as shown on the next page. Due Date: 18 Novembe, 014, 5:00 pm 3

4 Electicity and Magnetism: PHY Novembe, 014 a b c Rmax Rmax O Rmax θmax θ Section showing a disk at height We beak the sphee into disks. Each disk is at height ϵ[ R, R] fom the equatoial plane. Each disk compises ings at adius ϵ[0, R max = R ] fom the cente of the disk. We fist find the cuent due to a ing. this is given by i = ω ρ d d. This ing poduces a field at O, whose -component is given by µ 0 ω µ 3 d d + 3/ In class we have deived the axial field due to a cuent caying loop. Now the total field at O due to a disk at height will be given by, Let s fist solve the integal R max =0 µ 0 ωρ Rmax= R d =0 3 d = G. + 3/ 3 d + 3/ Let + = x d = dx d = dx. 3 d = x dx. Theefoe Due Date: 18 Novembe, 014, 5:00 pm 4

5 Electicity and Magnetism: PHY Novembe, 014 G = R x x 3/ R x 3/ dx = x dx = 1 [ R x 1/ dx = 1 x 1/ R 1/ x 1/ 1/ [ 1 = R + R 1 ] = R + R = R + R. since R max = R R R ] x 3/ dx Hence the magnetic field due to the entie evolving sphee is: µ 0 ωρ +R R + R R d = µ +R 0ωρ R + 0 R d [ = µ 0 ωρ R R R 3 R ] 0 R 0 ] = µ 0 ωρ [R + R 3 R = µ 0ωρR 3 4. A hollow cone like a paty hat has vetex angle θ, slant height L and suface chage density σ. It spins aound its symmety axis with angula fequency ω. What is the magnetic field at the tip? Due Date: 18 Novembe, 014, 5:00 pm 5

6 Electicity and Magnetism: PHY Novembe, θ θ l R l dl Chage on the dashed stip = σπrl dl Cuent i due to the stip = σωrl dl Field due to a small section inside the ing = µ 0 Field s -component due to a small section = σωl sin θ dl. irl dϕ = µ 0 il sin θ dϕ l l = µ 0il sin θ dϕ l Field s -component due to the ing at height l Total field B = µ 0σω sin 3 θ = µ 0il sin θ l = µ 0il sin θ l π ϕ=0 = µ 0σω sin 3 θ dl L 0 dϕ dl = µ 0σω sin 3 θl Due Date: 18 Novembe, 014, 5:00 pm 6

7 Electicity and Magnetism: PHY Novembe, How should the cuent density inside a thick cylindical wie depend on so that the magnetic field has constant magnitude inside the wie? I am looking fo a mathematical answe. R Assuming that the wie has adius R, conside a cicula Ampeian Path inside the wie at adius, whee < R, as shown in the diagam above. B.d l = µ 0 j.da ClosedP ath ShadedAea whee LHS = Bπ. Suppose j and da ae paallel, then the RHS becomes µ 0 j.d A = µ 0 jda = µ 0 π j dϕd ShadedAea ShadedAea = µ 0 π j d. =0 =0 ϕ=0 Now B at the adius is given by B = µ 0 =0 j d. If B is independent of, =0 j d must be popotional to, that is, =0 j d = K, whee K is a constant. By inspection we must have j = K, such that =0 j d = =0 Kd = K. Hence j = K/ o j = K/. Thus the cuent density must tape off as 1/ inside the conducto if the field inside the conducto emains unifom. Due Date: 18 Novembe, 014, 5:00 pm 7

Magnetic Field of a Wire

Magnetic Field of a Wire Magnetic Field of a Wie Fundamental Laws fo Calculating B-field Biot-Savat Law (long method, but woks always) Ampee s Law (high symmety) B-Field of a Staight Wie Fo a thin staight conducto caying cuent