TEST 3

1. (25 points) A curious high school physics student tries out a crude generator by laying a movable conducting rod on a U shaped conductor. In this setup, a uniform magnetic field is perpendicular to the area bounded by the U shaped conductor and moving rod. The U shaped conductor and moving rod create a closed conducting path for current. The length of the moving rod and width of U shaped conductor is L = 15.0 cm. The student pulls the rod to the right at speed v = 20.0 cm/s. The magnetic field has magnitude B = 0.800 T and points in.

(a) (10 points) What direction does the current flow in the closed conducting path? Indicate this direction by circling clockwise or counterclockwise and drawing and arrow on the diagram above.

(b) (10 points) What is the magnitude |ε| of the induced voltage (emf) between the ends of the rod?

(c) (5 points) What is the magnitude E of the electric field felt by the electrons in the rod? What is the direction of this field? Indicate the direction with a labeled arrow.

SOLUTION:
(a) Counter-clockwise since the expanding area  is like a field externally  increasing inward. Vector-B_ind is out to OPPOSE this change. 
(b) BLV= (0.800 T)*(0.15 m)*(0.20 m) = 0.0240 (V)
(c) E*L = change in voltage; thus E = BLV/L = B*V = 0.160 (V/m) where  Vector -E points up. 

---------------------------------------------------------------------------- 

2. ( 53 points) Below is a model for a pulsed laser used in science and medicine. The interior of the pulse can represented by an EM plane wave with wavelength 1062 nm. Note: 1 nm = 10 ^-9 m. The magnetic field oscillates vertically up and down and has a magnitude of 7.00x10 ^-9 T . At a given instant the magnetic field vector points down as shown in the diagram below. The downward arrow is labeled by the magnitude B of the magnetic field . As shown by the circle with crossbars, the wave’s propagation direction is into the page.

(a) (14 points) At the instant described by the diagram, what is the direction of the electric field ? Indicate this direction by drawing an arrow symbol on the diagram below.

(b) (14 points) What is the magnitude of the electric field?

(c) (5 points) What is the frequency f of this electromagnetic wave?
(d) (20 points) If the pulse lasts for 16x10 ^-12 seconds, then what is
the length (in meters) of the pulse?

SOLUTION:
(a) The electric field points horizontally  to the right. Point your right fingers in the electric field direction and WRAP your fingers into the B-vector, and you thumb points in , in the direction of the propagation.  
(b) E =c*B = (3.00x10⁸ m/s)*(7x10^-9T) = 2.1 V/m.
(c) f = c/lambda = c/wavelength = (3.00x10⁸ m/s)/(1062x10^-9m) =- 2.82x10¹⁴ Hz.
(d) length = c*t = ( 3.00x10⁸ m/s)*(16x10 ^-12 s) = 0.0048 m = 4.8 mm.
---------------------------------------------------------------------------------------

3. (38 points) A person stands in front of a mirror at an amusement park. In the diagram below, the person (the object) is shown and to the right is a mirror, not shown. The mirror shows an upright image (not shown) of the person that seems to be 4.0 meters behind the mirror. Assume the image is 3.5 times the person’s (object’s) height .

(a) (2 points) Is the image virtual or real? Explain.

(b) (28 points) What is the focal length of the mirror?
(c) (4 points) Is the mirror concave or convex ?

(d) (4 points) On the diagram above sketch the mirror on the right of the person (object). Draw the image by ray racing as we did in class. Ray trace by drawing two rays from the tip of the arrow, one parallel to the axis and the other intersecting the point where the mirror crosses the axis. Label the object distance d_o and image distance d_i.

SOLUTION:

 (a) Obviously the image is virtual because the image is behind the mirror; this is true in the case of a flat mirror or a curved mirror.

 (b)  1/f = 1/d_o  + 1/d_i  where we need d_o. Note d_i = -4.0 m and the magnification m = + 3.5 = -d_i/d_o.  Thus,  solve for d_o = -(-4.0 m)/(3.5) = 1.1429.   Thus 1/f = 1/(1.1429 m)   + (-1/4.0 m) or f =
(1.1429)*(4.0)/(4.0 - 1.1429)  = 1.6 (m). 
Concave since f > 0.

(d)  See class notes or check out figure 23.16 and COMPARE  with figure 23.14.
--------------------------------------------------------------------------------

4. (37 points) A beam of light is emitted in a pool of water (n = 1.33) from a depth of 61.0 cm. The light ray strikes the surface a horizontal distance x from the spot directly above the light source as shown in the diagram below. For air, the index of refraction is n_air =1. SHOW WORK.



(a) (19 points) Suppose the beam strikes the water – air surface a horizontal distance 112.5 cm from the spot directly above the light source. Will the light exit the water ? Answer yes or no. If yes, give the angle the beam makes with the normal after it exits the water into the air (n_air =1). Diagram the problem in the white spaces below.

(b) (18 points) Suppose the beam strikes the water – air surface a horizontal distance 45.7 cm from the spot directly above the light source. Will the light exit the water ? Answer yes or no. If yes, give the angle the beam makes with the normal after it exits the water into the air (n_air =1). Diagram the problem in the white spaces below.

SOLUTION:
Your main task should be finding the critical angle; after that procedure, everything else falls in place.
Sin θc  = (1/1.33) giving  = 48.75 degrees.  You can also beforehand
compute tan theta = x/61.0 cm; yo,  theta is the incident angle on the water side and if theta > θc, the light will NOT exit.
(a)  theta = tan ^-1 ( 112.5/61.0) = 61.5 degrees >  θc so light will not exit.  
(b)   theta = tan ^-1 (45.7/61.0) = 36.83 degrees <θc ,  so it exits and you can find the exit angle in the following way using Snell's law:

(1.33)*sin 36.38   = (1)*sin theta_f, where theta_f is the final exit angle. See #30, Ch. 23 for practice with this concept in preparation for the final examination .  In any event, the exit angle is theta_f = 52. 8 degrees.