Electromagnetism Answers to Problem Set 9 Spring 2006

Transcription

1 Electromagnetism Answers to Problem et 9 pring 006. Jackson Prob. 5.: Reformulate the Biot-avart law in terms of the solid angle subtended at the point of observation by the current-carrying circuit. B(r) = µ 0I dl (r r ) π = µ 0I dl π = µ 0I π dl Let V = dl then B(r) = µ 0I π V The ith component of V may be written V i = dl î where î is the unit vector along the ith axis. By virtue of toke s theorem this can be converted into a surface integral V i = da î r r n = da n r r î where is a surface bounded by the circuit and where the direction of the surface normal n is related to the sense of the current (l ) by the right-hand rule. The above equation can be rewritten as n V = da Therefore with Now B(r) = µ 0I W n W = da W = ( W ) W = da n (r r ) + = Ω + 0. da n δ(r r )

2 The second integral vanishes since r is on a surface bounding the circuit, which is away from the observation point r. The first integral is, as shown on page in Chap. of the text, the solid angle Ω subtended at the observation point by the circuit that bounds. Therefore, B(r) = µ 0I Ω Example: Consider a point of observation on the z axis above a circular loop of radius a in the xy plane that carries current I. The loop subtends a solid angle Ω = π( cos θ), where θ is the angle between the z axis and a line from the point of observation to the loop. Thus z Ω = π a + z and Therefore Ω = π a + z + z πa ẑ = (a + z ) / (a + z ) B(z) = µ 0I a ẑ, (a + z ) / / ẑ. confirming a result obtained in class directly from the Biot-avart law.. Jackson Prob 5.: Find B z inside a uniformly wound solenoid. Use result from Prob. 5. to write the contribution from a segment of the solenoid of length dz as db z = µ 0NIdz dω dz = µ oni dω. where Ω = π( cos θ) where θ is the angle that the line from the observation point to the ring at z makes with the axis. Integrate from end to end to find B z = µ oni Ω Ω = µ 0NI cos θ cos θ. In terms of the angles shown in the figure in the text this becomes. Jackson Prob 5.7: B z = µ 0NI cos θ + cos θ (a) As shown in class and in example with Prob. 5., the field of a single loop in the xy plane at a distance z from its center on the axis is B z = µ 0I a a + z /

3 (b) The field near the center of a Helmholtz pair is, therefore, B z = µ 0I a a + (z b/) + a / a + (z b/) / = µ 0Ia d + ( b a ) z d ( a 4 6b a + b 4) z 4 6 d 8 where d = b + 4a. 7 ( 5a 6 0b a 4 + 5b 4 a b 6) z 6 6 d + (c) ρ dependence of of field. On the axis, we may write B z = σ 0 + σ z +, where the coefficients σ k can be inferred from the above equation. With the aid of B = 0 we find that near the origin, (ρb ρ ) ρ = ρ B z z = σ ρ z + olving for B ρ (taking into account that B ρ = 0 on axis) we find that near the axis, B ρ (z, ρ) σ ρ z From B = 0, we may write olving for B z, we obtain B z (z, ρ) B z (z, 0) σ ρ B z ρ = B ρ z σ ρ = σ 0 + σ ) (z ρ + (d) From mathematica the asymptotic series for B z is B z == µ 0Ia z + ( b a ) z + 5 ( a 4 6b a + b 4) 6 z 4 7 ( 5a 6 0b a 4 + 5b 4 a b 6) 6 z 6 + which can be obtained from the power series by the replacement d z. (e) For a Helmholtz coil one sets b = a. With this choice the terms in the bracket for the small z expansion become 44z4 5a 4,,

4 Thus, to differ from uniformity by ɛ, the fractional distance must satisfy z/a (5/44) ɛ /4. For ɛ = 0 4 the limit is and for ɛ = 0 the limit is Below is a figure showing the variation of B z on the axis between two coils located at a = ±. B z z for a Helmholtz coil z 4. Jackson Prob 5.: Find the vector potential and magnetic induction for a uniformly charged sphere of radius a rotating about an axis with angular momentum ω. We orient ω along the z axis and let r lie in the xz plane. The vector r is used to locate a point on the sphere. The surface current density at r is K(r ) = ω r σ. The corresponding vector potential is J(r )d r K(r )da ω r da A(r) = µ 0 We write µ 0 = µ 0σ ω r = aω sin θ ( sin φ ˆx + cos φ ŷ) As in the example worked out in ec. 5.5 of the text, only the y component can contribute to the integral. Therefore, A y (r) = µ 0σa ω sin θ cos φ dω r + a ar cos γ, where cos γ is the angle between r and r. Expanding the denominator in a series of spherical harmonics we obtain A y (r) = µ 0σa ω r l < r l+ > l + Y lm(ˆr) dω sin θ cos φ Y lm (ˆr ) lm We first carry out the φ integral to find π dφ cos φ Y lm (θ, φ (l + )(l )! ) = π (l + )! 0 Noting that sin θ = P (cos θ ), we find sin θ P l (µ )dµ = 4 δ l 4 P l (cos θ ) (δ m δ m ).

5 Putting the previous two results together, we find dω sin θ cos φ Y lm (ˆr ) = 8π (δ m δ m )δ l The sum over lm above becomes lm Finally, r l < r l+ > = 6π 9 8π (Y (ˆr) Y (ˆr)) r < r = > sin θ r < r > In vector form, this becomes A φ (r) = µ 0σa 4 ω sin θ r = µ 0σa ω r sin θ r > a A(r) = µ 0σa 4 ω r r r > a = µ 0σa ω r The corresponding formulas for the magnetic induction B = A are B(r) = µ 0σa 4 ( ) (ω ˆr)ˆr ω r r > a = µ 0σa ω Note: A somewhat different (and simpler) solution to this problem is found in the text by Griffiths. He chooses coordinates with r be along the z axis and ω in the xz plane at an angle θ with the z axis. 5