A vertically oriented square loop of wire falls from a region where the
magnetic field **B** is horizontal, uniform and perpendicular to the plane of the
loop, into a region where the field is zero. Let the length of each side
be s and the diameter of the wire be d. The resistivity of the wire is ρ_{R}
and the density of the wire is ρ_{m}. If the loop reaches terminal
velocity while its upper segment is still in the magnetic field region, find an
expression for the terminal speed.

Solution:

- Concepts:

The induce emf ε = -∂flux/∂t, Ohms law, energy conservation. - Reasoning:

As the loop falls, the flux through the loop changes and an emf is induced in the loop. This causes a current to flow in the loop, and energy is dissipated in the form of heat. If the loop falls with constant velocity, then the rate at which the gravitational potential energy decreases must equal the rate at which heat is generated. - Details of the calculation:

|emf| = |dflux/dt| = Bsdy/dt, I = emf/R. Here dy/dt is the speed with which the loop is falling.

When terminal speed is reached: I^{2}R = emf^{2}/R = Fv_{terminal}= mgv_{terminal}.

The energy dissipated in the wire per unit time equals the gravitational potential energy lost per unit time.

(Bsv_{terminal})^{2}/R = mgv_{terminal}, v_{terminal}= mgR/(Bs)^{2}.

R = ρ_{R}16s/πd^{2}, m = ρ_{m}πd^{2}s, v_{terminal }= 16g ρ_{m}ρ_{R}/B^{2}.

A conducting circular loop made of wire of diameter d,
resistivity ρ, and mass density ρ_{m}
is falling from a great height h in a magnetic field with a component B_{z }= B_{0}(1 + kz),
where k is some constant. The loop of diameter D is always
parallel to the x-y plane. Disregard air resistance, and find the
terminal velocity of the loop.

Solution:

- Concepts:

The induce emf ε = -∂F/∂t, Ohms law, energy conservation. - Reasoning:

As the loop falls through the inhomogeneous magnetic field, the flux through the loop changes, and an emf is induced in the loop. This causes a current to flow in the loop, and energy is dissipated in the form of heat. If the loop falls with constant velocity, then the rate at which the gravitational potential energy decreases must equal the rate at which heat is generated. - Details of the calculation:

The gravitational force is in the -z direction,**F**_{g}= -mg**k**.

The magnetic flux through the loop is F = BA = B_{0}(1 + kz)πD^{2}/4.

The emf induced in the loop is ε = -∂F/∂t = -(B_{0}kπD^{2}/4)dz/dt.

The current flowing in the loop is I = |ε|/R.

The rate at which heat is generated is I^{2}R. The rate at which gravitational potential energy decreases is mg dz/dt

When dz/dt = -v_{t}, (v_{t}= terminal velocity), then I^{2}R = -mg dz/dt = mgv_{t}.

Then |ε|^{2}/R = mgv_{t}, [B_{0}^{2}k^{2}π^{2}D^{4}/(16R)]v_{t}^{2}= mgv_{t}, v_{t}= 16mgR/(B_{0}^{2}k^{2}π^{2}D^{4}).

m = ρ_{m}V = ρ_{m}(πd^{2}/4)πD. For a wire R = ρ length/area = ρ 4D/d^{2}.

v_{t}= 16ρ_{m}(πd^{2}/4)πDg(ρ 4D/d^{2})/(B_{0}^{2}k^{2}π^{2}D^{4}) = 16ρ_{m}gρ/(B_{0}^{2}k^{2}D^{2}).

A thin disk of ordinary metal with electrical conductivity σ has radius R and thickness d. It is held fixed in a
perpendicular magnetic field B(t)**k** = (B_{0} + αt)**k** where B_{0} and α
are positive constant quantities. In the following, neglect the self inductance
of the disk. See the sketch.

(a) Find the current density vector **j**(r) at
distance r from the axis of the disk.

(b) Determine the energy/time delivered to the disk by the
field. What becomes of this energy?

(c) Suppose that the constant
a becomes negative, i.e., that
α -->
—α. How do your results in
parts (a) and (b) change?

Solution:

- Concepts:

Faraday's law, Ohms law - Reasoning:

Divide the disk into concentric rings. The magnetic flux through the ring is changing with time. This flux change induces an emf. Since the ring has finite resistance, a current flows in the ring. - Details of the
calculation:

For a ring with inner radius r and outer radius r + dr we have

ε(r) = -∂F/∂t = induced emf, ∂F/∂t = (∂B/∂t)пr^{2}= αпr^{2}.

E = αпr^{2}/(2пr) = αr/2. If α is positive,**E**points into the –φ direction.

**j**= σ**E**, j(r) = σαr/2.

(b) The current flowing in the ring with radius r is j(r)*d*dr. The rate at which energy is delivered to the ring is ε(r)*j(r)*d*dr = (dпr^{3}σα^{2}/2)dr.

The rate at which energy is delivered to the disk is

(dпσα^{2}/2)∫_{0}^{R}r^{3}dr = dпσα^{2}R^{4}/8. This energy is converted into thermal energy.

(c)**E**and**j**now point into the φ direction. The power delivered to the disk is the same.

A uniform magnetic field **B** = B_{0}**k**
points in the z-direction. A particle with mass m and charge
q moves with kinetic energy E_{0} in the x-y plane in a circular orbit
centered at the origin as shown.

At t = 0 the magnetic field strength starts changing
slowly, so that at t = t_{1} it is **B** = B_{1}**k**.
Neglect radiation.

(a) What is the radius of the orbit R_{0} of the particle for t < 0 in
terms of B_{0}, and E_{0}?

(b) Assuming that the radius R of the orbit of the particle does not change
appreciably while the particle completes one revolution, what is the kinetic
energy E_{1} of the particle at time t_{1} in terms of B_{0},
B_{1}, and E_{0}?

Now assume that the magnetic field stays
constant (**B** = B_{0}**k**), but that the particle is subject to
a drag force **F**_{d} = -m**v**/τ, where τ is a constant. At
time t = 0 the position and velocity of the particle are **R**_{0} =
(0, R_{0}, 0) and **v**_{0} = (v_{0}, 0, 0).

(c) Write Newton's equations of motion for the velocity components v_{x},
v_{y}, and v_{z}.

(d) Construct an equation of motion for z = v_{x} + iv_{y}, and
solve it.

(e) Find expressions for v_{x}(t), v_{y}(t), x(t), and y(t).
Describe the trajectory of the particle in words.

Solution:

- Concepts:

Faraday's law, induced emf, damped oscillations - Reasoning:

A changing magnetic flux induces an emf which changes the kinetic energy of the particle. - Details of the calculation:

(a) R_{0}= mv/(qB_{0}) = (2mE_{0})^{½}/(qB_{0})

(b) Assume that the radius R of the orbit of the particle does not change appreciably while the particle completes one revolution. Then the particle's kinetic energy changes by

ΔE = q(dB/dt)πR^{2}over one revolution.

The time for one gyration is T = 2πR/v,

so dE/dt = ΔE/T = (qvR/2)(dB/dt) = (qmv^{2}R/(2mv))(dB/dt) = (1/B)(dB/dt)E.

(We used R = mv/(qB), qR/(mv) = 1/B, ½mv^{2}= E.)

dE/E = dB/B, ln(E_{1}) = ln(B_{1}E_{0}/B_{0}), E_{1}= B_{1}E_{0}/B_{0}.

(c) F_{x}= mdv_{x}/dt = qv_{y}B_{0}– mv_{x}/τ, F_{y}= mdv_{y}/dt = -qv_{x}B_{0 }– mv_{y}/τ, F_{z}= 0.

(d) dz/dt = -i(qB_{0}/m)(v_{x}+ iv_{y}) – (v_{x}+ iv_{y})/τ = -i(qB_{0}/m)z - (1/τ) z

= -i[(qB_{0}/m)z - (i/τ)]z = gz,

where g = -i[(qB_{0}/m)z - (i/τ).

z(t) = z_{0}exp(gt) = z_{0}exp(-t/τ) exp(-iωt), with ω = qB_{0}/m.

(e) Re(z(t)) = v_{x}(t) = v_{0}exp(-t/τ) cos(ωt), Im(z(t)) = v_{y}(t) = -v_{0}exp(-t/τ) sin(ωt).

x(t) = ∫_{0}^{t}v_{0}exp(-t'/τ) cos(ωt') dt'

= v_{0}[exp(-t/τ)[- cos(ωt)/τ + ω sin(ωt)] + 1/τ]/[1/τ^{2}+ ω^{2}]

y(t) = -∫_{0}^{t}v_{0}exp(-t'/τ) sin(ωt') dt' + R_{0}

= -v_{0}[exp(t/τ)[ -sin(ωt)/τ – ω cos(ωt)] + ω] /[1/τ^{2}+ ω^{2}] + R_{0}

As t --> infinity, x(t) --> x_{∞}= (v_{0}/τ)/[1/τ^{2}+ ω^{2}], y(t) --> y_{∞}= -v_{0}ω/[1/τ^{2}+ ω^{2}] + R_{0 }The particle spirals into the point (x_{∞}, y_{∞}).

A thin metallic square frame of mass m, electrical
resistance R, and side a is rotating about an axis perpendicular to a uniform
magnetic field B as shown in the figure. Initially the square frame
rotates with a frequency ω_{0}.

(a) Determine the average energy loss per cycle due to Joule heating.

(b) Determine the time it takes for the frequency of the rotation to slow down
to 1/e of its initial value. (Assume that the fractional change in the frame's
rotation frequency per cycle is small.)

Solution:

- Concepts:

The induce emf e = -∂F/∂t, Ohms law, energy conservation - Reasoning:

A changing magnetic flux produces an emf in the frame. The emf causes a current to flow which produces Joule heating. Kinetic energy is converted to thermal energy. - Details of the calculation:

(a) (SI units)

The magnetic flux through the frame is Ba^{2}cosθ. The angle is changing, dθ/dt = ω.

The induced emf due to this changing flux is -Ba^{2}sinθ ω.

The current flowing in the frame is given by I = Ba^{2}sinθ ω/R (neglecting the self inductance of the loop).

The amount of kinetic energy converted into thermal energy per unit time is P = B^{2}a^{4}sin^{2}θ ω^{2}/R.

Energy loss per cycle (assuming ω changes negligibly):

∫_{0}^{T }Pdt = ∫_{0}^{2π }(P/ω)dθ = πω B^{2}a^{4}/R = ΔE

Initial average energy loss per cycle due to Joule heating: πω_{0}B^{2}a^{4}/R.

<P> = ΔE/T = B^{2}a^{4}ω^{2}/2R(b) <P> = B

^{2}a^{4}ω^{2}/2R = -dE/dt

E = ½Iω^{2}, = ma^{2}ω^{2}/12, ω^{2}= 12E/ma^{2}, dE/dt = -E 6B^{2}a^{2}/(Rm)

E = E_{0}exp[-6B^{2}a^{2}/(Rm)t), t_{(1/e)}= Rm/(6B^{2}a^{2})