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Electricity and Magnetism: PHY-204. 11 November, 2014 Collaborative ASSIGNMENT Assignment 3: Sources of magnetic fields Solution 1. (a) A conductor in the shape of a square loop of edge length l m carries a current I as shown in the figure below. Calculate the magnitude and direction of the magnetic field at the center of the square. (b)What If ? If this conductor is formed into a single circular turn and carries the same current, what is the value of the magnetic field at the center? Answer z dz y 90+θ r x z θ l/2 Field inwards. Due Date: 18 November, 2014, 5:00 pm 1 Electricity and Magnetism: PHY-204. 11 November, 2014 |dBz | = µ0 I|d⃗ℓ × r̂| ( ( )) 4π z 2 + ℓ 2 2 = µ0 I dz sin( π2 + θ) ( )) ( 4π z 2 + ℓ 2 2 µ0 I = ( ( )2 ) cos θ dz. 4π z 2 + 2ℓ Now z ℓ/2 = tan θ, |dBz | = = ∴ |Bz | = = = ⇒ z = 2ℓ tan θ ⇒ dz = 2ℓ sec2 θ dθ. ( )2 µ0 I ℓ sec2 θ dθ cos θ ( ℓ )2 2 2 4π 2 (sec θ) µ0 I cos θ dθ 2πℓ ( ) ∫ + tan−1 z ℓ/2 µ0 I ( ) cos θ dθ z 2πℓ θ=− tan−1 ℓ/2 +π/4 µ0 I sin θ−π/4 2πℓ ( ) µ0 I 1 µ0 I · √ =√ · (Fields are additive in this case) 1 2πℓ 2 + √2 2πℓ Furthermore a wire of length 4ℓ implies a circle of radius 2ℓ . π 2l π Field at its center is, µ0 B = I 4π ∫ 2π ( ϕ=0 ) 2ℓ dϕ ( 2ℓ )2 π π µ0 I π · 2π = 4π 2ℓ µ0 πI = · 4ℓ Due Date: 18 November, 2014, 5:00 pm 2 Electricity and Magnetism: PHY-204. 11 November, 2014 2. A conductor consists of a circular loop of radius R and two straight, long sections as shown in the figure below. The wire lies in the plane of the paper and carries a current I. Find the vectorial expression for the magnetic field at the center of the loop. Answer From infinite wire, B1 = From circle, B2 = µ0 I , 2R µ0 I , 2πR into the plane of the paper. also into the plane of the paper. (see the previous question). Hence B= µ0 I µ0 I + , 2πR 2R into the plane of the paper. 3. A sphere of radius R has a uniform volume charge density ρ. Determine the magnetic field at the center of the sphere when it rotates as a rigid object with angular speed ω about an axis through its center, as shown on the next page. Answer Due Date: 18 November, 2014, 5:00 pm 3 Electricity and Magnetism: PHY-204. 11 November, 2014 (a) (b) (c) Rmax(z) r z Rmax r O Rmax(z) θmax θ Section showing a disk at height z We break the sphere into disks. Each disk is at height zϵ[−R, R] from the equatorial √ plane. Each disk comprises rings at radius rϵ[0, Rmax (z) = R2 − z 2 ] from the center of the disk. We first find the current due to a ring. this is given by i = ω ρ r dr dz. This ring produces a field at O, whose z-component is given by µ0 ω µ r3 dr dz · 2 (z 2 + r2 )3/2 (In class we have derived the axial field due to a current carrying loop). Now the total field at O due to a disk at height z will be given by, µ0 ωρ dz 2 Let’s first solve the integral ∫ √ Rmax = R2 −z 2 r=0 ∫ Rmax r=0 r3 dr (z 2 +r2 )3/2 = G. Let z 2 + r2 = x ⇒ 2rdr = dx ⇒ rdr = Due Date: 18 November, 2014, 5:00 pm r3 dr · (z 2 + r2 )3/2 dx . 2 r3 dr = (x − z 2 ) dx . Therefore 2 4 Electricity and Magnetism: PHY-204. ∫ R2 G = z2 R2 ∫ = z2 1 = 2 [∫ 11 November, 2014 (x − z 2 ) dx x3/2 2 (since Rmax = R2 − z 2 ) x−3/2 (x − z 2 ) dx 2 R2 x −1/2 z2 z2 dx − 2 ∫ ] R2 x −3/2 dx z2 2 R −1/2 R 2 z2 x 1 x1/2 − = 2 1/2 z2 2 (−1/2) z2 [ ] 1 2 1 = (R − z) + z − R z z2 = R−z+ −z R z2 = R+ − 2z. R Hence the magnetic field due to the entire revolving sphere is: ) ( ∫ µ0 ωρ +R z2 − 2z dz R+ 2 R −R ) ( ∫ 2µ0 ωρ +R z2 = − 2z dz R+ 2 R 0 ] [ R 1 3 R 2 2 R z − z 0 = µ0 ωρ Rz 0 + 3R 0 2 [ ] 2 R = µ0 ωρ R2 + − R2 3 µ0 ωρR2 = · 3 4. A hollow cone (like a party hat) has vertex angle 2θ, slant height L and surface charge density σ. It spins around its symmetry axis with angular frequency ω. What is the magnetic field at the tip? Answer Due Date: 18 November, 2014, 5:00 pm 5 Electricity and Magnetism: PHY-204. 11 November, 2014 90-θ θ l z R(l ) dl Charge on the dashed strip = σ2πR(ℓ) dℓ Current i due to the strip = σωR(ℓ) dℓ = σωℓ sin θ dℓ. Field due to a small section inside the ring = µ0 iR(ℓ) dϕ µ0 iℓ sin θ dϕ = · 2 4π ℓ 4π ℓ2 Field’s z-component due to a small section = µ0 iℓ sin2 θ dϕ · 4πℓ2 Field’s z-component due to the ring at height ℓ ∫ µ0 iℓ sin2 θ 2π = dϕ 4πℓ2 ϕ=0 µ0 iℓ sin2 θ 2ℓ2 µ0 σω sin3 θ dℓ = · 2 = Total field µ0 σω sin3 θ B= 2 Due Date: 18 November, 2014, 5:00 pm ∫ L dℓ = 0 µ0 σω sin3 θL · 2 6 Electricity and Magnetism: PHY-204. 11 November, 2014 5. How should the current density inside a thick cylindrical wire depend on r so that the magnetic field has constant magnitude inside the wire? I am looking for a mathematical answer. Answer R r Assuming that the wire has radius R, consider a circular Amperian Path inside the wire at radius r, where r < R, as shown in the diagram above. ∫ ∫ ⃗ ⃗l = µ0 ⃗ ⃗j(⃗r ).dA B.d ClosedP ath ShadedArea ⃗ are parallel, then the RHS becomes where LHS = B(2πr). Suppose ⃗j(r) and dA ∫ µ0 ∫ ∫r ∫2π ⃗ = µ0 ⃗j(⃗r ).dA ShadedArea j(r)dA = µ0 ShadedArea ∫r j(r′ )r′ dϕdr′ r′ =0 ϕ=0 j(r′ )r′ dr′ . = µ0 2π r′ =0 Now B at the radius r is given by µ0 B= r ∫r ∫r j(r′ )r′ dr′ . r′ =0 If B is independent of r, r′ =0 j(r′ )r′ dr′ must be proportional to r, that is, ∫r j(r′ )r′ dr′ = Kr, where K is a constant. r′ =0 ∫r ∫r By inspection we must have j(r′ )r′ = K, such that r′ =0 j(r′ )r′ dr′ = r′ =0 Kdr′ = Kr. Hence j(r′ ) = K/r′ or j(r) = K/r. Thus the current density must taper off as 1/r inside the conductor if the field inside the conductor remains uniform. Due Date: 18 November, 2014, 5:00 pm 7