In the LHC, there are actually two circular tubes, side-by-side, that protons travel through. The protons go one way in one tube, and the other way in the other tube. ( And then eventually the two paths are made to cross to smash the protons together.) The diagram below shows a short segment of the two tubes of radius r and separation d (black), with coils of wire on them (red) that carry a current I to generate the magnetic fields. The wires are essentially long straight wires that run right along the sides of the tubes for a length L, and then wrap over the top to the other side. The drawing just shows one red line, but this is supposed to represent N loops of wire forming a bundle with that shape. The wires are coated with an insulator so that current flows along the length of wire only, and not from one wire to another in the bundle.

(a) If we want the protons in the upper tube in the figure to go
counterclockwise, and the protons in the lower tube to go clockwise, find the
direction of current flow in the upper and lower coils (clockwise or
counterclockwise for each).

(b) Near the midpoint of the straight segment of the wire, we can neglect the
magnetic fields from the ends, and consider the wires to be just infinite
straight, parallel wires. Find the magnetic field (magnitude and direction) at
the center of each tube, near the midpoint of the straight sections. Your answer
should be in terms of r, d, I, and N.

(c) Given the numbers B = 8 T, r = 50 mm, d = 250 mm, and N = 80, evaluate the
current I in the wires.

(d) Ignoring the semicircles at the ends, calculate the force (magnitude and
direction) on the bundle of wires on one side of a pipe due to the bundle of
wires on the other side of the pipe. First find a symbolic equation in
terms of N, I, L, and r, then plug in the numbers above and L = 15 m.

Solution:

- Concepts:

Ampere's law, the force on a current-carrying wire - Reasoning:

Since L >> r, we use the formulas for infinitely-long wires. - Details of the calculation:

(a) The force on the protons must point towards the center of the ring. The magnetic field at the center of the upper tube must point into the page and the magnetic field at the center of the lower tube must point into the page. The currents must flow as shown below.

(b) Ampere's law:(upper tube) = [μ

B_{0}NI/(2π)](1/r + 1/r +1/(d - r) - 1/(d + r))**⛒**= [μ

_{0}NI/(2π)](2/r +2r/(d^{2}- r^{2}))**⛒.**(lower tube) = [μ

B_{0}NI/(2π)](2/r + 2r/(d^{2}- r^{2}))**☉**.

(c) I = [2πB/(μ_{0}N)](2/r + 2r/(d^{2}- r^{2}))^{-1}= 1.2*10^{4}A

(d) F = ILB = NIL[μ_{0}NL/(2π)]/(2r) = [μ_{0}N^{2}I^{2}L/(2π)]/(2r).

The wires repel each other.

F = 1.38*10^{7}N.

Consider a particle of charge q and mass m in the presence of a constant,
uniform magnetic field B = B_{0 }**k**, and of a uniform electric
field of amplitude E_{0}, rotating with

frequency ω in the (x,y) plane, either in the clockwise or in counterclockwise
direction.

Let **E** = E_{0}cos(ωt) **i** ± E_{0}sin(ωt) **
j**.

(a) Write down the equation of motion for the particle and solve for
the Cartesian velocity components v_{i}(t) in terms of B_{0}, E_{0},
and ω if ω ≠ ω_{c} = qB_{0}/m (the cyclotron
frequency).

Show that, if ω = ω_{c}, a resonance is observed for the appropriate sign of ω.

Hint: Let ζ = v_{x} + iv_{y}, and solve for ζ(t).

(b) Solve for the Cartesian velocity components v_{i}(t) at
resonance.

(c) Now assume the presence of a frictional force **f** = -mγ**v**, where
**v** is the velocity of the particle. Find
the general solution for ζ(t) for clockwise rotation of the electric field, and
find the steady state solution (t >> 0) when ω = ω_{c}.

Solution:

- Concepts:

The Lorentz force,**F**= q(**E**+**v**×**B)**= md**v**/dt - Reasoning:

dv_{z}/dt = 0, v_{z}= constant. We have to solve coupled differential equations for v_{x}and v_{y}using the hint, ζ = v_{x}+ iv_{y}. - Details of the calculation:

(a) dv_{x}/dt = qv_{y}B_{0}/m + (qE_{0}/m)cos(ωt),

dv_{y}/dt = -qv_{x}B_{0}/m ± (qE_{0}/m)sin(ωt),

Let ζ = v_{x}+ iv_{y},

dζ/dt = -iqB_{0}ζ/m + (qE_{0}/m)exp(±iωt), or

dζ/dt + iω_{c}ζ = (qE_{0}/m)exp(±iωt),

We find the solution to the inhomogeneous equation dζ/dt + iω_{c}ζ = (qE_{0}/m)exp(±iωt)

and add the solution of the homogeneous equation dζ/dt + iω_{c}ζ = 0.

inhomogeneous solution:

Try ζ = ζ_{0}exp(±iωt), ±iωζ_{0}+ iω_{c}ζ_{0}= (qE_{0}/m) ζ_{0}= -i(qE_{0}/m)/(ω_{c}± ω).

homogeneous solution:

ζ = Aexp(-iω_{c}t), A = arbitrary complex constant = |A|exp(iφ).

general solution:

ζ = Aexp(-iω_{c}t) - i(qE_{0}/m)exp(±iωt)/(ω_{c}± ω).

v_{x}(t) = |A|cos(ω_{c}t + φ) ± (qE_{0}/m) sin(ωt)/(ω_{c}± ω).

v_{y}(t) = -|A|sin(ω_{c}t + φ) - (qE_{0}/m) cos(ωt)/(ω_{c}± ω).

We have a resonance (the expression is undefined) at ω = ω_{c}and the field rotates clockwise.

i.e.**E**= E_{0}cos(ωt)**i**- E_{0}sin(ωt)**j**.

(b) At resonance dζ/dt + iω_{c}ζ = (qE_{0}/m)exp(-iω_{c}t),

Try a solution with a time-dependent amplitude, ζ = C(t)exp(-iω_{c}t).

dC/dt - Ciω_{c}+ iω_{c}C = qE_{0}/m. C(t) = qE_{0}t/m + C_{0}. C_{0}= arbitrary complex constant.

v = (qE_{0}t/m + C_{0})exp(-iω_{c}t).

v_{x}(t) = (qE_{0}t/m) cos(ω_{c}t) + |C_{0}|cos(ω_{c}t + φ).

v_{x}(t) = -(qE_{0}t/m) sin(ω_{c}t) - |C_{0}|sin(ω_{c}t + φ).

(c) The equations of motion now are

dv_{x}/dt = qv_{y}B_{0}/m - γv_{x}+ (qE_{0}/m)cos(ωt),

dv_{y}/dt = -qv_{x}B_{0}/m - γv_{y}- (qE_{0}/m)sin(ωt),

dζ/dt + iω_{c}ζ + γζ = (qE_{0}/m)exp(-iωt).

inhomogeneous solution:

Try ζ = ζ_{0}exp(-iωt), ±iωζ_{0}+ iω_{c}ζ_{0}+ γζ_{0}= (qE_{0}/m) ζ_{0}= -i(qE_{0}/m)/(ω_{c}- ω -iγ).

homogeneous solution:

ζ = Aexp(-iω_{c}t -γt), A = arbitrary complex constant = |A|exp(iφ).

general solution:

ζ = Aexp(-iω_{c}t -γt) - i(qE_{0}/m)exp(-iωt)/(ω_{c}- ω -iγ).

The first term decays exponentially.

The steady state solution for ζ at resonance is ζ = (qE_{0}/(γm)) exp(-iω_{c}t).

Refer to the figure. One end of a conducting rod rotates with angular
velocity ω in a circle of radius a making contact with a horizontal, conducting
ring of the same radius. The other end of the rod is fixed. Stationary
conducting wires connect the fixed end of the rod (A) and a fixed point on the
ring (C) to either end of a resistance R. A uniform vertical magnetic field
**B** passes through the ring.

(a) Find the current I flowing through the resistor and the rate at which
heat is generated in the resistor.

(b) What is the sign of the current, if positive I corresponds to flow in the
direction of the arrow in the figure?

(c) What torque must be applied to the rod to maintain its rotation at the
constant angular rate ω?

What is the rate at which mechanical work must be done?

Solution:

- Concepts:

Motional emf - Reasoning:

The conducting rod is moving in a plane perpendicular to**B**. - Details of the calculation:

(a) Speed of the rod as a function of the distance r from the origin: v(r) = ωr.

Force on an electron: F_{e}= q_{e}vB = q_{e}Bωr. (towards point A)

Work done per unit charge: emf = Bω∫_{0}^{a}rdr = Bωa^{2}/2.

Assume that the resistance of the conducting rod and the wires is negligible.

I = emf/R = Bωa^{2}/(2R) is the current flowing through the resistor.

P_{e}= I^{2}R = B^{2}ω^{2}a^{4}/(4R) is the rate heat is generated.

(b) The sign of the current is positive.

(c) Force on a section dr of the current carrying rod: dF = IdrB (direction clockwise in the figure).

An external force of equal magnitude and opposite direction is needed to maintain the constant angular speed.

dτ = rdF = r IBdr, τ = IB∫_{0}^{a}rdr = IBa^{2}/2 = B^{2}ωa^{4}/(4R).

The rate at which mechanical work is done is P_{m}= τω = B^{2}ω^{2}a^{4}/(4R) = P_{e}.

A^{ }square loop made of wire with negligible resistance is placed^{
}on a horizontal frictionless table as shown (top view). The^{ }
mass of the loop is m and the length of^{ }each side is b. A
non-uniform vertical magnetic field exists^{ }in the region; its
magnitude is given by the formula^{ }B = B_{0} (1 + kx), where B_{0}
and k are known constants.^{ }

The loop is^{ }given a quick push with an initial velocity v along^{
}the x-axis as shown. The loop stops after a time^{ }interval t.
Find the self-inductance L of the loop.^{ }

Solution:

- Concepts:

Motional emf - Reasoning:

The loop has no resistance. The motional emf must be canceled by the induced emf due to the self inductance of the loop. - Details of the calculation:

Let v(t) be positive, if it points into the positive x-direction and let I(t) be positive if it flows clockwise in the loop. When the loop moves with positive velocity v(t), the motional emf = v(t)bB(x '+ b) - v(t)bB(x') = v(t)b^{2}B_{0}k causes a current I(t) to flow clockwise. Since the loop has no resistance, Kirchhoff's loop rule requires vb^{2}B_{0}k - LdI/dt = 0. The motional emf is canceled by the induced emf due to the self inductance of the loop. We have

(1) vb^{2}B_{0}k = LdI/dt.

The magnetic force on the current carrying loop is

F = mdv/dt = -IbB(x' + b) + IbB(x') = -Ib^{2}B_{0}k.

The acceleration points in the negative x-direction if the direction of current flow is clockwise. We have

(2) dv/dt = -Ib^{2}B_{0}k/m

We can differentiate (1) with respect to t and obtain

(3) dv/dt = (L/(b^{2}B_{0}k))d^{2}I/dt^{2}.

Combining (2) and (3) we obtain

(4) d^{2}I/dt^{2}= -[(b^{2}B_{0}k)^{2}/(mL)] I.

The solution to this second-order differential equation satisfying I(t = 0) = 0 is

I(t) = I_{max}sin(ωt), with ω^{2}= (b^{2}B_{0}k)^{2}/(mL).

The velocity v is proportional to dI/dt, so v(t) = v_{max}cos(ωt).

The loop oscillates about its starting position along the x-direction, v(t) = 0 for the first time for ωt = π/2. This yields L = (1/m)(2b^{2}B_{0}kt/π)^{2}in terms of the given parameters.

A current I flows in the circuit shown below. Calculate the magnetic field at P as a function of the current I and the distances a and b. Segments BC and AD are arcs of concentric circles. Segments AB and DC are straight-line segments.

Solution:

- Concepts:

The Biot-Savart law - Reasoning:

The Biot-Savart law is used to find the field of an arbitrary current distribution. - Details of the calculation:

**B**(**r**) = (μ_{0}/(4π))∫ I d**l'**x (**r**-**r**')/|**r**-**r**'|^{3 }Let the point P lie at the origin of the coordinate system whose z-axis points out of the page.

For the field produced by the current flowing along the circular arc with radius b we have(0) = (μ

B_{0}/4π)I(2π/3)b^{2}/b^{3}**k**= (μ_{0}/6)I/b**k**,

where**k**is the unit vector pointing out of the page.

For the field produced by the current flowing along the circular arc with radius b + a we have(0) = -(μ

B_{0}/6)I/(b + a)**k**.

The field produced by the straight-line segments at the origin is zero.

The field at the origin therefore is

**B**(0) = (μ_{0}/6)I (a/(b(b +a ))**k**.