Physics 505 Fall 2005 Homework Assignment #8 Solutions

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1 Physics 55 Fall 5 Homework Assignment #8 Solutions Textbook problems: Ch. 5: 5., 5.4, 5.7, A circular current loop of radius a carrying a current I lies in the x-y plane with its center at the origin. a) Show that the only nonvanishing component of the vector potential is A φ (ρ, z) = µ Ia where ρ < (ρ > ) is the smaller (larger) of a and ρ. dk cos kz I (kρ < )K (kρ > ) The vector potential may be obtained by A( x ) = µ J( x ) 4 x x d3 x where (for a circular current loop) J( x ) = Iδ(z )δ(ρ a) ˆφ in cylindrical coordinates. Note that to obtain the cylindrical components of A( x ) we have to be careful to convert the basis vector ˆφ at the point x to components at x. (This is because the basic vectors depend on position.) A bit of geometry gives ˆφ = ˆρ sin(φ φ ) + ˆφ cos(φ φ ) Or, alternatively, we may choose the point x to lie at φ =, so that ˆφ = ŷ and ˆρ = ˆx. Then it is straightforward to see that ˆφ = ŷ cos φ ˆx sin φ = ˆφ cos φ ˆρ sin φ. Using symmetry, we can see that only the ˆφ component of A is nonvanishing. The integral expression for the vector potential is then A( x) = µ I δ(z )δ(ρ a)ˆρ sin(φ φ ) + ˆφ cos(φ φ ) 4 x x ρ dρ dφ dz = µ Ia ˆρ sin(φ φ ) + ˆφ () cos(φ φ ) 4 x x dφ where the integrand in the second line is to be evaluated at z = and ρ = a. We now use the cylindrical Green s function expressed as x x = 4 dk cosk(z z ) I (kρ < )K (kρ > ) + cosm(φ φ )I m (kρ < )K m (kρ > ) m=

2 Note that the integral over φ picks out the m = term in the sum. Furthermore, the ˆρ component drops out because sin(φ φ ) is orthogonal to cos(φ φ ), a result that could have been obtained by symmetry. We end up with A( x ) = µ Ia 4 = µ Ia 4 ˆφ ˆφ b) Show that an alternative expression for A φ is A φ (ρ, z) = µ Ia dk cos(kz)i (kρ < )K (kρ > ) dk cos(kz)i (kρ < )K (kρ > ) dk e k z J (ka)j (kρ) To obtain the alternative expression, we use the alternate form of the Greens function x x = dk e k(z > z < ) J (kρ)j (kρ ) + m= cosm(φ φ )J m (kρ)j m (kρ ) Since, for z =, we have z > z < = z, it is clear that when we stick this into () we end up with A( x ) = µ Ia ˆφ dk e k z J (kρ)j (ka) c) Write down integral expressions for the components of magnetic induction, using the expressions of parts a) and b). Evaluate explicitly the components of B on the z axis by performing the necessary integrations. Since B = A and the only non-vanishing component of A is A φ, we end up with B ρ = z A φ, B z = ρ ρ(ρa φ ) The z derivative is straightforward. For the ρ derivative, on the other hand, we may use the Bessel equation identity d dz X (z) + z X (z) = X (z) where X m denotes either J m, N m, I m or K m. This gives, in particular ρ ρρx (kρ) = kx (kρ)

3 Hence, for the expression of a) we find and B ρ = µ Ia B z = µ Ia dk k sin(kz)i (kρ < )K (kρ > ) dk k cos(kz) { } I (kρ)k (ka) I (ka)k (kρ) where the top line holds for ρ < a, while the bottom line holds for ρ > a. Similarly, the vector potential of b) yields the magnetic induction B ρ = µ Ia sgn(z) dk ke k z J (kρ)j (ka) and B z = µ Ia dk ke k z J (kρ)j (ka) The z axis corresponds to ρ =. In this case, it is easy to see that B ρ = (a result demanded by symmetry) follows from the result that either J () = or I () =. For the B z component, we take the representation of part b). Noting that J () =, we end up with B z (ρ = ) = µ Ia = µ Ia = dk ke k z J (ka) a (z + a ) 3/ µ Ia (z + a ) 3/ which agrees with the elementary result for a current loop on axis. This integral was performed by noting that it is a Laplace transform L{tJ (at)}, which in turn is the derivative d/ds of the transform L{J (at)}. The Laplace transform of a Bessel function can be looked up, with the result L{J n (at)} = a n ( s + a s) n / s + a. 5.4 A long, hollow, right circular cylinder of inner (outer) radius a (b), and of relative permeability µ r, is placed in a region of initially uniform magnetic-flux density B at right angles to the field. Find the flux density at all points in space, and sketch the logarithm of the ratio of the magnitudes of B on the cylinder axis to B as a function of log µ r for a /b =.5,.. Neglect end effects. For a long cylinder (neglecting end effects) we may think of this as a twodimensional problem. Since there are no current sources, we use a magnetic scalar

4 potential Φ M which must be harmonic in two dimensions. Since H = Φ M, we orient the uniform magnetic field H along the +x axis and write ( H ρ + α ρ ) cos φ, ρ Φ M (ρ, φ) = (βρ + γ ρ ) cos φ, a δρ cos φ, > b < ρ < b ρ < a Of course, the general harmonic expansion would be of the form (A m ρ m + B m ρ m ) cos mφ + (C m ρ m + D m ρ m ) sin mφ. However here we have already used the shortcut that all matching conditions for m lead to homogeneous equations admitting only a trivial (zero) solution. The magnetostatic boundary conditions demand that H φ and B ρ are continuous at both ρ = a and ρ = b. The magnetic field (and magnetic induction) components are and H φ = ( H + α ρ ρ ) sin φ, ρ > b φφ M = (β + γ ρ ) sin φ, a < ρ < b δ sin φ, ρ, a µ ( H α ρ ) cos φ, ρ > b B ρ = µ ρ Φ M = µ(β γ ρ ) cos φ, a < ρ < b µ δ cos φ, ρ < a The resulting matching conditions at a and b are H + α b = β + γ b, H α b = µ r (β γ b ) () β + γ a = δ, β γ a = µ r δ where µ r = µ/µ. These equations may be solved to yield where α = (µ r µ r )(b a )H β = ( + µ r )H γ = ( µ r )a H δ = 4 H ( a ) = (+µ r )(+µ r )+( µ r )( µ r ) = b µ r (µ r + ) (µ r ) ( a b The magnetic scalar potential is then given by () with the above values of the coefficients. We see that the magnetic induction for ρ < a is uniform, pointed )

5 along the same direction as B. The other two regions contain a dipole field in addition a uniform component. Since H = Φ M = δˆx for ρ < a, the ratio of B on axis (ρ = ) to B is given by B = 4 4 = B ( + µ r )( + µ r ) + ( µ r )( µ r )(a/b) This may be plotted as follows B/ B.8 ( a / b ) = ( a / b ) =. log µ r 5.7 A current distribution J( x ) exists in a medium of unit relative permeability adjacent to a semi-infinite slab of material having relative permeability µ r and filling the halfspace, z <. a) Show that for z > the magnetic induction can be calculated by replacing the medium of permeability µ r by an image current distribution, J, with components, ( µr µ r + ) J x (x, y, z), ( ) µr J y (x, y, z), µ r + ( ) µr J z (x, y, z) µ r + We will end up solving parts a) and b) simultaneously. We start, however, by defining the reflection (Parity) operator P : z z so that P : (x, y, z) (x, y, z) On the right (z > ), we assume the magnetic induction is generated by both the original current J (contained entirely on the right) and an image current J (contained entirely on the left). Thus B R ( x ) = µ 4 ( J( x ) + J ( x )) ( x x ) x x 3 d 3 x By changing variables z z in the J term, we may restrict this volume integral to z > B R ( x ) = µ ( J( x ) ( x x ) J 4 x x 3 + ) (P x ) ( x P x ) x P x 3 d 3 x (3) z >

6 On the left (z < ), we assume the magnetic induction is generated by a current of the same form as the original J, but with possibly modified strength (because of the change of permeability). Given a modified current λj and permeability µ, we write B L ( x ) = µλ 4 z > J( x ) ( x x ) x x 3 d 3 x (4) Our aim is now to match the left and right magnetic field and magnetic induction. More precisely, at z =, both H x and H y (the parallel components) must be continuous, and B z (the perpendicular component) must also be continuous. To perform this matching, we first note that the norms x x and x P x are identical at z =. (The are both equal to (x x ) + (y y ) + z.) Thus all denominators are the same, and we deduce that the numerators of (3) and (4) must be matched as appropriate. For B z, we have (J x + J x)(y y ) (J y + J y )(x x ) = µ r λ(j x (y y ) J y (x x )) where any component of J is understood to have argument P x. For H x and H y matching, we find (J y J y )z (J z + J z )(x x ) = λ( J y z J z (x x )) (J z + J z )(x x ) + (J x J x)z = λ(j z (x x ) + J x z ) Since these equations hold for all values of (x, y), they separate into λj y = J y J y λj z = J z + J z λj z = J z + J z λj x = J x J x These equations may be solved to yield µ r λj x = J x + J x µ r λj y = J y + J y J x = ( λ)j x, J y = ( λ)j y, J z = ( λ)j z provided µ r λ = λ, or λ = /(µ r + ). This may be given in a more concise form using the reflection operator J ( x ) = ( λ)p J(P x ) = µ r µ r + P J(P x ) b) Show that for z < the magnetic induction appears to be due to a current distribution µ r /(µ r + ) J in a medium of unit relative permeability. From the expression (4) for B L, the magnetic induction appears to be due to a current λ J = /(µ r + ) J in a medium of permeability µ. This is equivalent

7 to having a current distribution µ r /(µ r + ) J in a medium of unit relative permeability. 5.9 A magnetically hard material is in the shape of a right circular cylinder of length L and radius a. The cylinder has a permanent magnetization M, uniform throughout its volume and parallel to its axis. a) Determine the magnetic field H and magnetic induction B at all points on the axis of the cylinder, both inside and outside. We use a magnetic scalar potential and the expression Φ M = 4 V M( x ) x x d 3 x + 4 S ˆn M( x ) x x Orienting the cylinder along the z axis, we take a uniform magnetization M = M ẑ. In this case the volume integral drops out, and the surface integral only picks up contributions on the endcaps. Thus Φ M = M 4 top x x da bottom da x x da where top and bottom denote z = ±L/, and the integrals are restricted to ρ < a. On axis (ρ = ) we have simply Φ M (z) = M 4 = M 4 = M ( ) ρ + (z L/) ρ dρ dφ ρ + (z + L/) a ( ) ρ + (z L/) dρ ρ + (z + L/) a + (z L/) a + (z + L/) z L/ + z + L/ On axis, the field can only point in the z direction. It is given by H z = z Φ M = M z L/ a + (z L/) z + L/ a + (z + L/) sgn(z L/) + sgn(z + L/) Note that the last two terms cancel when z > L/, but add up to inside the magnet. Thus we may write H z = M z L/ a + (z L/) z + L/ a + (z + L/) + Θ(L/ z )

8 where Θ(ξ) denotes the unit step function, Θ = for ξ > (and otherwise). The magnetic induction is obtained by rewriting the relation H = B/µ M as B = µ ( H + M ). Since the magnetization is only nonzero inside the magnet ie M z = M Θ(L/ z ), the addition H + M simply removes the step function term. We find B z = µ (H z + M z ) = µ M z L/ a + (z L/) z + L/ a + (z + L/) b) Plot the ratios B/µ M and H/M on the axis as functions of z for L/a = 5. The z component of the magnetic field looks like H / M z/ L while the z component of the magnetic induction looks like B / µ M z/ L Note that B z is continuous, while H z jumps at the ends of the magnet. This jump may be thought of as arising from effective magnetic surface charge.