TECHNICAL PHYSICS II   TEST NO. 3   SPRING 2008

     1. Two long, straight, parallel wires 20 cm apart each carry a current
	of 2.4 A but in opposite directions.  A third parallel wire is 20 cm
	from each of these two wires (that is, making an equilateral triangle)
	and carries a current of 1.2 A as shown in the figure below.

                                         (.)  1.2 A out of the page



           2.4 A out of the page   (.)         (x)  2.4 A into the page

	What force per unit length is exerted (magnitude and direction) on
	the third wire?

	The force due to the left wire on the upper wire is attractive,
	directed 60° below the horizontal to the left (or 60° below
	the negative x axis for a conventional x-y coordinate system).  The
	force due to the right wire on the upper wire equals that of the left
	wire in magnitude, but is repulsive and is directed 60° above the
	negative x axis.  By symmetry, the vertical (or y component) forces
	cancel and the resultant force must be to the left (along the negative
	x axis).  The resultant force is therefore just the sum of the two
	equal x-component forces.

	F/L = 2[&mu:oII' / (2πr)]cos60°

	where I = 2.4 A, I' = 1.2 A, and r = 0.20 m.  Therefore,

	F/L = 2.88 × 10-6 N/m, directed to the left.


     2. Magnetic field in a coaxial cable:  The central wire of a coaxial
	cable carries a current of I.  Surrounding the wire (and coaxial with
	it) is an outer conductor, a hollow cylinder with an inner radius = a
	and an outer radius = b.  It is also carrying a current of I but in
	the opposite direction as that in the wire.  Find the magnitude of the
	magnetic field as a function of the radius r, assuming the central
	wire to be very small (approx. zero radius).  That is, find the
	formula for B versus r for a > r > 0, b > r > a, and r > b.  Use
	Ampere's law, assuming the current in the outer conductor is uni-
	formly distributed over its cross section.  (Remember:  As a check
	B should be continuous at r = a and r = b.)

	a > r > 0:

	The enclosed current for this case is just I, so

	B = μoI / (2πr).

	b > r > a:

	The enclosed current is that of the central wire, = I, plus the
	current (flowing in opposite direction, hence negative) in that part
	of the hollow cylinder where r < b. The current flowing in the
	cylinder is

	I(πr2 - πa2) / (πb2 - πa2)

	= I(r2 - a2) / (b2 - a2).

	The total enclosed current is

	Iencl = I - I(r2 - a2) / (b2 - a2)

	= I(b2 - r2) / (b2 - a2).

	The magnetic field is therefore

	B = [μoI / (2πr)](b2 - r2) / (b2 - a2).

	r > b:

	The enclosed current is zero so B = 0.

	If you check the first two results for r = a and the second two for
	r = b, you will find they agree.


     3. Magnetic storms generated by solar flares can cause power outages
	on earth.  Consider a square conducting loop 25.0 km on a side lying
	horizontally on the surface of the earth.  (This will be a model of a
	power grid.)  The earth's magnetic field at this grid has a magnitude
	of 5.5E-5 T and is oriented at an angle of 45 degrees to the vertical.
	What would be the average emf generated around this loop if the
	magnetic field increased by 10% in 0.250 s?

	The magnetic flux is

	ΦB = BAcos45° = (5.5 × 10-5 T)(25000 m)2cos45° = 2.43 × 104 Wb.

	The change in the flux is 10% of this or 2.43 × 103 Wb.

	This change takes place in 0.25 s, so the average emf is

	ΔΦB / Δt = 9700 V.  (Wb/s =  V)


     4. A long, straight solenoid has a length of 50.0 cm, a radius of
	5.00 cm, and 1200 turns.  What is the self-inductance of the
	solenoid?

	The self-inductance of a coil is L = NΦB / I.  For a
	long solenoid, the magnetic field can be thought of as approximately
	uniform and equal to

	B = μonI = μo(N / l)I,

	where l = length of the solenoid.  The flux in the coil is therefore

	ΦB = BA = μo(N / l)Iπr2.

	Combining this with the equation for L you get

	L = μoN2πr2 / l = 28.4 mH.


     5. A possible means of space flight is to place a perfectly reflecting
	aluminized sheet into orbit around the earth and then use the light
	from the sun to push this "solar sail".  Suppose a square sail, 1.00
	km on a side with a mass of 5000 kg is placed in orbit facing the sun.
	What force is exerted on the sail by sunlight (assuming a solar
	intensity of 1340 watts per square meter due to the fact the sail is
	the same distance from the sun as Earth) and, ignoring the effect of
	gravity, what would be the sail's acceleration?

	The formula for radiation pressure on a perfectly reflective surface
	is

	prad = 2I / C,

	where I is the intensity of the light.  Hence,

	prad = 2(1340 W/m2) / (3 × 108 m/s) = 8.93 × 10-6 N/m2.

	Since force = pressure times area, the force is

	F = (8.93 × 10-6 N/m2)(1000 m)2 = 8.93 N.

	a = F / m = 0.0018 m/s2.