Questions: 1 2 3 4

PHYS 2420 Magnetic Fields in Materials Solutions

1. A permanent magnet has the shape of a right circular cylinder of length L. If the magnetization M is uniform and has the direction of the cylinder axis, find the surface and volume current densities.






We are told that the magnetization is M = Mk, where M is a constant.

The surface current density is given by J_SM = M ´ n, where n is the outward looking normal to the surface, and M is evaluated at the point in question. We have three surfaces here. For surface A₁, n₁ = −k, that is the surface normal points to the left. For surface A₂ the surface normal points right, n₂ = +k. For A₃, n₃ = r, where r is the cylindrical radial unit vector. Thus the charge densities are

J_SM1 = M × n₁|_x=0 = Mk × k|_x=o = 0,

J_SM2 = M × n₂|_x=-L = Mk ´ -k|_x=L = 0 ,

and

J_SM3 = M × n₃|_r=R = Mk ´ r|_r=R .

There are no surface currents on the ends of the cylinder since the magnetization is parallel to the normal.

To interpret the meaning of this cross product, lets draw a 2D end-on view of the end of the cylinder.



Note that k ´ r is tangential to the surface cylinder and in the counterclockwise sense. In Cartesian coordinates,

,

so that

k ´ r = θ ,

and

J_SM3 = M × n₃|_r=R = Mθ .

The volume current density is given by

J_M = Ñ ´ M .

Since the magnetization is constant, its curl is zero. There are no volume currents.



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2. Find the distribution of magnetization currents corresponding to a uniformly magnetized sphere with magnetization M.


For convenience let M = Mz, where M is a constant. The diagram below shows the situation.



The volume current density is given by

J_M = Ñ ´ M .

Since the magnetization is constant, its curl is zero. There are no volume currents.

The surface current density is given by J_SM = M ´ n, where n is the outward looking normal to the surface, and M is evaluated at the surface in question. We have one surface here. For a sphere n = r, where r is the cylindrical radial unit vector. Thus the charge densities are

J_SM = M × n₁|_r=R = Mk ´ r|_r=R .

To go further we need to express r in terms of Cartesian coordinates, .

Thus we find



We have used the conversions between Cartesian and spherical coordinates, x = Rsinθcosφand y = Rsinθsinφ, in the above derivation. The unit vector φ is always tangential to the sphere in the manner of lines of latitude on the earth. The magnitude of the surface currents is maximum at the equator and zero at the poles of the sphere.
 



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3. A sphere of magnetic material of radius R is placed at the origin of coordinates. The magnetization is given by M = (ax²+b)i, where a and b are constants. Determine the magnetization current densities.






The volume current density is given by

J_M = Ñ ´ M .

Since the magnetization is only in the x direction, M = (M_x, 0, 0) this reduces to

.

So there are no volume currents.

The surface current density is given by J_SM = M ´ n, where n is the outward looking normal to the surface, and M is evaluated at the surface in question. We have one surface here. For a sphere n = r, where r is the cylindrical radial unit vector. Thus the charge densities are

J_SM = M × n₁|_r=R = Mi ´ r|_r=R .

To go further we need to express r in terms of Cartesian coordinates,

.

Thus we find



We have used the conversions between Cartesian and spherical coordinates, y = Rsinθsinφand z = Rcosθ, in the above derivation. The surface currents are tangential to the sphere in yz slices, with the maximum currents at x = 0 and no currents at x = <± R.



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4. A conducting cylinder is coated with a magnetic material with inner radius a, outer radius b, and permeability μ . The cylinder carries a uniform current I. Determine the magnetic field H, the magnetic flux density B, the magnetization M, and the magnetization currents.






When dealing with magnetic materials, we start with the magnetic field H. Since we have cylindrical symmetry, we can use the integral form of Ampere's Law

ò_CH•dl = òj(r)dS ,

where ò j(r)dS = I_enclosed, the enclosed free current. Free current is the move of free charge not the induced magnetic surface and volume currents. Symmetry demands that the magnetic field be tangential, H = H(r)θ . In cylindrical coordinates, the unit vector θ indicates the tangential direction, the directions we might refer to as clockwise or counterclockwise about the central axis of the cylinder. To exploit the symmetry we choose path C to be a circle. Thus at r = R

ò_CH•dl= H(R)2πR .

As well, ò j(r)dS reduces to 2pò j(r)rdr. Thus Ampere's Law reduces to

H(R)2πR = 2pò_r=0^r=R j(r)rdr .

We know that the current I is uniformly distributed through the inner cylinder. The current density is



Thus the problem is naturally broken into two regions, inside the metal cylinder and outside. Remember that the magnetic field H is unaffected by the presence of magnetic materials and magnetization currents.

1. 0 £ R £ a

H(R)2πR	= 2pòr=0r=R j(r)rdr
	= 2pòr=0r=R (I/πa2) rdr
	= 2π (I/πa2)[ ½r2| r=0r=R]
	= 2π R2 (I/2πa2)

Thus we have

H(R)= IR/2πa² .

2. a £ R

H(R)2πR	= 2pòr=0r=R j(r)rdr
	= 2π {òr=0r=a (I/πa2) rdr + òr=0r=a (0) rdr}
	= 2π (I/πa2)[ ½r2| r=0r=a]
	= 2π (I/2π)

Thus we have

H(R) = I/2πR .

The magnetic flux density B is affected by the presence of magnetic materials and is related to H by

B = μ₀H ,

where μ₀ is the permeability of the material where you wish to determine B. Hence B = Bθ, the flux density is tangential since H is tangential. We need to consider three regions since there are three materials here, inside the conductor, inside the magnetic sheath, and outside in the air (vacuum).

1. 0 £ R £ a

In a conductor, μ = μ₀ which is the permeability of free space.

So in this region,

B(R) = μ₀IR/2πa² .

2. a < R £ b

In the sheath, we are given μ so

B(R) = μI/2πR .

3. b < R

   We are in a vacuum so

   B(R) = μ₀I/2πR .

The magnetization M is related to B and H by

M = B/μ₀ – H .

Since B and H are tangential, M = M(R)θ . Since B varies in the three regions, so does M.

1. 0 £ R £ a

   M = (μ₀IR/2πa²)/μ₀ –  IR/2πa² = 0 .

   This is right since we have no induced dipoles in the conductor.

2. a < R £ b

   M(R) = (μI/2πR)/μ₀ – I/2πR = (μ/μ₀ – 1)(I/2πR) .

3. b < R
M(R) = (μ₀I/2πR)/μ₀ – I/2πR = 0 .

This is right because there is nothing to magnetize in a vacuum.

Since we know the magnetization everywhere, we can find the volume and surface current densities that are created by it. In particular, since the magnetization is zero outside the magnetic sheath, the currents can only occur on and in this sheath.

The surface current density is given by J_SM = M ´ n, where n is the outward looking normal to the surface, and M is evaluated at the point in question. We found that the magnetization is M = M(R)θ, with M(R) given above. For the sheath, we have four surfaces, each flat end of the cylindrical sheath and the inside and outside of the sheath. For front end of the cylinder n₁ = −k, that is the surface normal points to the left. For the back end the normal points right, n₂ = +k. For the inside of the cylindrical sheath n₃ = −r, where r is the cylindrical radial unit vector. For the outside n₄ = +r. Thus the charge densities on the flat ends are

J_SM1 = M × n₁|_r = M(R)q ´ –k|_r = –M(r)r ,

and

J_SM2 = M × n₂|_r = M(R)q ´ k|_r = +M(r)r ,

where we have used the crossproduct identities for unit cylindrical vectors.

These currents are radially inwards at the front end with a density M(a) at the inside rim and M(b) at the outside rim.

The charge densities for the inside and outside are

J_SM3 = M × n₃|_R=a = M(R)q ´ –r|_R=a = +M(a)k .

and

J_SM4 = M × n₃|_R=b = M(R)q ´ r|_R=b = −M(b)k .

These currents run the length of the cylinder. On the inside they travel parallel to the real current I. On the outside they travel in the antiparallel to I. The diagram below tries to indicate the surface current directions. Note that these directions assume μ and therefore M are positive. If the material is diamagnetic, μ is negative and the currents are opposite to those shown.



The volume current density is given by

J_M = Ñ ´ M .

Due to the cylindrical symmetry of this problem, it is best to use the cylindrical representation of the curl. Recalling that we derived

M = (M_r, M_θ , M_z) = (0, , 0) ,

we find that

JM	= Ñ ´ M
	=
	=
	=
	= 0

There are no volume currents.

Questions?mike.coombes@kwantlen.ca