COLLEGE PHYSICS II   TEST NO. 3   SPRING 2008

     1. A coil 15.0 cm in diameter and wound with ten turns of wire is
	placed with the plane of the coil at right angles to a magnetic field
	of 0.0500 T.  What emf is induced in the coil if (a) the field goes
	to zero in 0.0350 s and (b) the field is reversed (goes from 0.0500 T
	in one direction to 0.0500 T in the other direction) in 0.0350 s?

	The magnetic flux in the coil goes from a certain amount to zero,
	which is a negative change, but you are only interested in the
	magnitude of the change in these Faraday law problems.  The
	flux is originally,

	ΦM = NBA = (10)(0.050 T)π(0.075 m)2 = 0.00884 Wb.

	Since it goes to zero, the magnitude of the change of the flux is also
	0.00884 Wb.  The magnitude of the average emf generated around 
	he coil is equal to the magnitude of flux change divided by the
	time elapsed during the flux change.  Therefore,

	emfav = (0.00884 Wb) / (0.0350 s) = 0.252 V.

	For (b) note that the field changes from 0.00884 Wb to 0 to 0.00884 Wb
	in the opposite direction.  This means the change in the flux is
	double that of (a), or 0.0177 Wb, and since the time of the
	change is the same, the emf must be double,

	emfav = 0.505 V.


     2. A sinusoidal oscillator with a frequency of 400 Hz and an effective
	voltage of 180 V is put in series with an inductor with self-
	inductance of 250 mH.  What is the effective current in the circuit,
	assuming the resistance is negligible?

	First find the reactance of the inductor,

	XL = 2πfL = 2π(400 Hz)(0.25 H) = 628 Ω.

	The current is given by

	I = V / XL = (180 V) / (628 Ω) = 0.286 A


     3. A laser emits 1.5E+18 photons every second, producing a beam with
	500 mW of power (that is, 500 millijoules per second).  What is the
	the energy per photon?  What is the frequency of the light?  (You
	should get a large number for the frequency, so don't necessarily
	panic at your answer.)

	Since there are 1.5 × 1018 photons emitted every
	second and 0.50 J of energy tranmitted in the beam every second, the
	energy per photon must be

	E = (0.50 J) / (1.5 × 1018) = 3.33 × 10-19 J.

	The frequency is given by

	f = E / h = (3.33 × 10-19 J) / (6.626 × 10-34 J/Hz) = 5.03 × 1014 Hz.


     4. A beam of light impinges on the top surface of a 4.00-cm thick
	parallel glass (n = 1.75) plate at an angle of 30 degrees to the
	normal.  How long is the actual path through the glass?

                 |
	_________|_______________________
                 |\
            4 cm |-\  d
                 |A \
	_________|___\___________________
                 |

	Kind of a funky diagram, but it indicates the 4 cm thickness of the
	glass, the angle of transmission A, and the path of the light beam
	in the glass, d.  (The 30° angle of incidence can't be shown with
	just keyboard symbols.)  Snell's law says angle A is

	A = sin-1[sin(30°) / 1.75] = 16.6°.

	d is the hypotenuse of the triangle and 5 cm is the side adjacent to
	angle A, so

	d = (4.00 cm) / cos(16.6°) = 4.17 cm.