Problem 1:
A circular loop of wire with radius a = 1 cm
and center at the origin is bent, so half lies in the y-z plane and
half in the x-y plane. A current I = 2 A flows in the wire.
(a) What is the magnetic moment of this loop?
(b) What is the magnetic field at (x, y, z) = (3, 4, 0) meters from the
origin?
Solution:
 | Concepts, principles, relations that apply to the
problem: The magnetic moment of a current loop, the principle of superposition, the dipole field |
| Why do they apply? |
We can view the loop as a superposition of loop 1 and loop 2 as shown. We find m by adding the moments of loop 1 and loop 2, m = m1 + m2.
| How do they apply? (a) m1= (-i)Ipa2/2, m2= (-k)Ipa2/2, m= (-i-k)Ipa2/2, m = Ipa2/21/2. (SI units) |
| Details of the calculation: The distance from the origin to the point where we want to know the field is much greater that the dimensions of the loop. The magnetic field is therefore the dipole field.
m×r = -3Ipa2/2,
(m×r)r = (-9Ipa2/2)i
+ (-6Ipa2)j, |
Problem 2:
A
thin rectangular sheet of composite material has dimensions L ´
w
with L >> w. The sheet carries a steady surface current that circulates as
shown.
A surface current density K
= dI/dy = constant
flows in the x direction over a width D. At the ends of
the rectangle
(x = ± L/2) a high conductance “short
circuit” completes the current path.
(a)
Find the magnetic moment m.
(b)
Sketch the magnitude of m as a function of the width D
of the current.
Solution:
| Concepts, principles, relations that apply to the
problem: The magnetic moment |
| Why do they apply? We are asked to find the magnetic moment of a current distribution. |
| How do they apply? (a)  Consider a loop as shown above. For this filamentary loop dm = k(Kdy)2Ly.  .(b)  |
| Details of the calculation: None |
Problem 3:
(a) Find the magnetic field B of a rotating spherical shell with
uniform surface charge density s.
(b) Find the magnetic field B of a uniformly magnetized
sphere.
Solution:
| Concepts, principles, relations that apply to the
problem: Boundary conditions, the uniqueness theorem, Magnetization M, magnetization current densities |
| Why do they apply? We can solve for the magnetic field of a rotating spherical shell with uniform surface charge density s, by treating the problem as a boundary value problem. The magnetization of a uniformly magnetized sphere results in a magnetization surface current density like the surface current density of a rotating spherical shell with uniform surface charge density s. |
| How do they apply? (a) Let the radius of the sphere be R, its center be located at the origin, and let the sphere rotate with angular velocity w = wk. The rotating surface charge density produces a surface current density, km = sw´R, km = swRsinq. For r >> R we will have a dipole potential,  or
Af = (m0/4p)msinq/r2.To find the dipole moment m, we consider the rotating sphere to be a stack of current loops. For a current loop at q we have dm = IA = RswsinqRdq pR2sin2q = R4swpsin3qdq. Integrating dm from q = 0 to q = p we find m = (4/3)R4swp, m = mk. [ňsin3xdx = -(1/3)cosx (sin2x+2)] For r >> R we have Af = (m0/3)R4swsinq/r2. We may write km = (3m/(4pR3)sinq. |
| Details of the calculation: ,
 ,
(SI units).Assume that A(r) will only has f component. Assume that the 1/r2 term is the only term in the expansion of A in powers of 1/rn outside of the sphere. Since A must be continuous at r = R and A must be finite at the origin, assume that Af = Csinqr = (m0/3)Rswsinqr inside the sphere. If this solution fulfills the boundary conditions,  ,
 we have the only solution. B = Ń´A, Bin_r = (1/)rsinq))¶(sinqAf_in)/¶q = (2m0/3)Rswcosq, = (m0/4p)2mcosq/R3. Bin_q = -(1/r)¶(rAf_in)/¶r = -(2m0/3)Rswsinq = -(m0/4p)2msinq/R3 Bout_r = (1/(rsinq))¶(sinqAf_out)/¶q = (2m0/3)R4swcosq/r3 = (m0/4p)2mcosq/r3 Bout_q = -(1/r)¶(rAf_out)/¶r = (m0/3)R4swsinq/r3 =(m0/4p)msinq/r3 Now check that the boundary conditions are satisfied. At r = R the radial component of B is continuous. Bout_q - Bin_q = m0R4swsinq/R3 = m0km. The boundary conditions are satisfied and we have the only solution. The field outside the sphere is a pure dipole field,  and the field inside the sphere is constant, B(r) = k(m0/4p)2m/R3. (b) For a uniformly magnetized sphere with radius R and M = Mk we have m = (4/3)pR3M. The magnetization current densities are jm = 0, km = M ´ R/R, km = Msinq . We may write km = (3m/(4pR3)sinq. Expressed in terms of the magnetic moment m the uniformly magnetized sphere has the same surface current density as the rotating spherical shell with uniform surface charge density s. The same currents produce the same fields. The field outside the uniformly magnetized sphere is a pure dipole field, and the field inside the sphere is constant, B(r) = k (m0/4p)2m/R3 = (m0/4p)(8p/3)M. |
Problem 4:
A small spherical cavity of radius a is made in a permanent magnet of uniform magnetization M. Find B and H at the center of the cavity.
Solution:
| Concepts, principles, relations that apply to the
problem: The principle of superposition |
| Why do they apply? Maxwell's equations are linear equations. The principle of superposition holds. We may write Bcavity = Bmaterial - Buniformly magnetized sphere. |
| How do they apply? For a uniformly magnetized sphere we have: Bout = pure dipole field, Bin = uniform, in the direction of M. Let M = Mk, m = (4/3)pa3M. Bout = (m0/4p)[3(m×r)r/r5 - m/r3]. Bout_q = (m0/4p)(m/r3)sinq, Bout_r = (m0/4p)(2m/r3)cosq. Bin = Bink. B is continuous at r = ak. Bin = (m0/4p)(2m/a3)k = (m0/4p)(8/3)pM. We therefore have Bcavity = Bmaterial - (m0/4p)(8/3)pM. Bmaterial depends on the shape of the magnetized material. Hcavity = Bcavity/m0. |
| Details of the calculation: None |
A short cylinder (length l and radius a) of iron is magnetized along the axis of the cylinder. Calculate H and B on the axis, both inside and outside.
Solution:
| Concepts, principles, relations that apply to the
problem: Magnetization M, magnetization current densities  the
field of a solenoid |
| Why do they apply? Magnetization always results in magnetization current densities. Uniform magnetization results in surface current densities. The uniform magnetization of a cylinder results in a surface current density like that of a solenoid. |
| How do they apply? For a solenoid we have Bz = (m0In/2)(cosq1 - cosq2). (We choose the symmetry axis to be the z-axis,)  (I*n) is the current per unit length, i.e. the surface current density. For the magnetized cylinder we therefore have on the axis Bz = (m0km/2)(cosq1 - cosq2).  ,
Bz = (m0M/2)(cosq1
- cosq2).If the center of the cylinder is at the origin then cosq1 = (z + l/2)/(a2 + (z + l/2)2)1/2, cosq2 = (z - l/2)/(a2 + (z - l/2)2)1/2. H = B/m0 outside. H = B/m0 - M inside. H = k[(M/2)(cosq1 - cosq2) - M] inside. At z = l/2 + e (e --> 0) we have H = k(M/2)cosq1, H points into the positive z-direction. At z = l/2 - e (e --> 0) we have H = k[(M/2)cosq1 - M], H points into the negative z-direction. H reverses direction at the boundary |
| Details of the calculation: None |
Problem 6:
A toroidal iron ring with permeability m = 1000 m0 is
wound with a uniform coil of 10 turns per centimeter, carrying a current of 5
amperes.
(a) Compute the magnitude of the B and H
fields inside the iron.
(b) The current is turned off, but B remains
unchanged due to permanent magnetization of the iron. What is H?
(c) The ring is cut and opened into a straight bar. Describe the resulting
B and H fields qualitatively.
Solution:
| Concepts, principles, relations that apply to the
problem: Ampere's law for H,  |
| Why do they apply? The problem has enough symmetry to let us calculate H from Ampere's law of H alone. Iron is not a lih material, but since m = 1000 m0 is given we assume that for the field strength B given in the problem we may write B = mH. |
| How do they apply? (a) n = 1000 turns/m, I = 5A. Consider a ring of radius r, R1 < r < R2.  2prH = In2pR1, H = InR1/r. The direction of H is the f direction. If the ring is very narrow, then R1 » R2 » R, and H = In. B = mH. B = 1000m0*5000A = 5*106*4p*10-7T = 2pT. |
| Details of the calculation: (b) Ampere's law for H yields H = 0, B = m0M. (c) We now have a bar magnet with uniform magnetization. See the previous problem. |
Problem 7:
A coil of N turns is wrapped around an iron ring of radius d and cross section A (d >> A1/2). Assume a constant permeability m >> 1 for the iron.
(a) What is the magnetic flux
as a function of
current I?
(b) If a gap of width b, (b2 << A) is cut in the
ring, what is the flux for the same current I?
(c) What is the field energy in the iron and in the gap?
(d) With such a gap in the ring, what is the self inductance?
Solution:
| Concepts, principles, relations that apply to the
problem: Ampere's law for H, ,
magnetic field energy, self inductance |
| Why do they apply? The problem has enough symmetry to let us calculate H from Ampere's law of H alone. Iron is not a lih material, but we assume that for the field strength B given in the problem we may write B = mH. After a small gap has been cut into the ring we use the boundary conditions for B to find B and H in the gap. We use the energy density u = (1/2)B×H to find the stored energy. |
| How do they apply? (a) H and B point into the f direction, H = NI/(2pr), B = mNI/(2pr).  If d >> A1/2, then B inside the coil is nearly constant, B = mNI/(2pd). Flux ,
f = mN2IA/(2pd)
µ I. |
| Details of the calculation: (b) If b2 << A, then Biron » Bgap = m0Hgap, because the normal component of B is continuous. Hiron(2pd - b) + Hgapb = NI, (B/m)(2pd - b) + (B/m0)b = NI, from Ampere's law for H. B = mm0NI/[m0(2pd - b) + mb], f = mm0N2IA/[m0(2pd - b) + mb]. (c) The energy density u = (1/2)B×H = (1/2m)B2. The field energy in the iron is Uiron = A(2pd - b)(m/2)(m0NI/[m0(2pd - b) + mb])2. The field energy in the gap is Ugap = Ab(m0/2)(mNI/[m0(2pd - b) + mb])2. Uiron + Ugap = (1/2)mm0N2I2A/[m0(2pd - b) + mb]. (d) f = LI or U = (1/2)LI2 yield L = mm0N2A/[m0(2pd - b) + mb]. |
Problem 8:
A current I is flowing in the metal strip of width L and thickness d shown above. The metal contains ne free electrons per unit volume. A magnetic field B penetrates the strip as shown. In terms of I, B, ne, and L, find the potential difference between side 1 and side 2.
Solution:
| Concepts, principles, relations that apply to the
problem: The Hall effect |
| Why do they apply? The current in the metal strip shown above is flowing because a power supply establishes an electric field E in the strip. The conduction electrons move with an average drift velocity vd opposite to the direction of E. In the uniform field B the magnetic force on each electron is F = -ev´B. It causes the electrons to drift towards side 1 of the strip, leaving excess positive charge on side 2. These separated charges produce the electric field EH. The electrons will stop drifting sideways, when the electric force -eEH cancels the magnetic force -ev´B. The resulting potential difference across the strip is D V = V2 - V2 = EHL = vdBL.This is known as the Hall potential. If positive charges were moving in the strip, they would drift towards side 1 and the sign of the Hall potential would change. |
| How do they apply? The magnitude of the current I moving through a wire is given by I = |r-|<v->A = neevdA, where ne is the number of free electrons per unit volume.
The current I equals the number of electrons that pass any point along the wire
per second times the unit of charge e. <v-> = vd is the drift speed of
the electrons. It is their average speed, as they move along the wire. |
| Details of the calculation: None |
Problem 9:
At the interface between one linear magnetic material and another the magnetic field lines bend. Show that tanq2/tanq1 = m2/m1, assuming there are no free currents at the boundary.
Solution:
| Concepts, principles, relations that apply to the
problem: Boundary conditions for B and H |
| Why do they apply? No free currents are at the interface. The normal component of B and the tangential component of H therefore have to be continuous at the boundary. |
| How do they apply? (a) Assume B1 makes an angle q1 with the normal to the interface in the medium with m1. B1x/B1y = tanq1.  By is continuous across the
boundary. Hx is continuous across the boundary. In the
medium with m1 we have H1x =
B1x/m1. In the medium
with m2 we have H2x = B2x/m2.
H1x = H2x. B1xm2/m1
= B2x. |
| Details of the calculation: None |
Problem 10:
At saturation, when nearly all of the atoms have their magnetic moments
aligned, the magnetic field in a sample of iron can be 2 T. If each
electron contributes a magnetic moment of one Bohr magneton, how many electrons
per atom contribute to the saturated field of the iron?
(Iron: ~8.5*1028 atoms/m3)
Solution:
| Concepts, principles, relations that apply to the
problem: The saturation magnetization Msat, B = m0(H + M). |
| Why do they apply? Msat = Nmk, for a magnetic field pointing in the z direction. Here m is the magnitude of the magnetic moment of each atom and N is the number of atoms per unit volume. A saturation B = m0Msat. |
| How do they apply? The magnetization M is defined as the magnetic dipole moment per unit volume. M = 8.5*1028 atoms/m3*n*9.27*10-24 Am2. Here n is the number of electrons per atom that contribute to the saturated field of the iron and 9.27*10-24 Am2 is the Bohr magneton in SI units. 2T = 4p*10-7(Tm/A)*8.5*1028 m-3*n*9.27*10-24 Am2, n = 2. |
| Details of the calculation: None |