QUIZ 10  , CHAPTER  29     PROBLEMS :  1, 2, 4(motional emf), 5, 6, 8, 9, 10, 11, 29 (motional emf),  generators
(motional emf): 36, 39, 40; 54 (Change in  magnetic flux produces an electric field even when there is no wire loop. DISCUSSION FOR 54: Find the electric field first, then find the | electric force|  = Q*E, where E is the magnitude of the  electric field.  )

TRANSFORMERS: 46, 47, 48.

Above,  PROBLEMS TAGGED "motional emf" mean  flux changes are due to the motion of the loop and/or the change in effective area of the loop. Generators, which convert mechanical  into electrical energy,  are based on this type of flux change. 


The other problems involve the loop having a stationary size or orientation and thus  the  flux changes are  due to the variation of the magnetic field with time.

FINALLY, TRANSFORMERS MERELY MAGNIFY THE EMF, much like a mechanical lever or hydraulic system  using Pascal's Law can magnify the force.  Note  even though there is an emf  or force amplification/reduction, energy is still conserved, since for example in the lever case,  f*D = F*d , where f  is for a small force taken through a large distance D and F  is for a larger  force taken through a smaller distance d. We find for transformers, energy conservation is reflected in  the relation: Vp * IP  =   Vs.*Is  .
Discussions  posted later. 

Simulation: Click here for the motion of an electron in a uniform electric field.

1. The flux changes uniformly.  Let the field point into the page initially and outward finally. That means the induced current will be clockwise   and the induced magnetic field inward. |emf| =  (area)*|change in B/change in t| =
(final flux  -  initial flux)/ change in t. The area is not needed since the initial and final fluxes are given.
2. Let the North pole of the external magnet be facing into the page; that means the external field  through the loop is also into the page. The magnitude of the filed is increasing. Thus,  the current in the loop will be counterclockwise  and the induced magnetic field  thus points out of the page. FOR EXTRA UNDERSTANDING RECALL THE FARADAY'S LAB.
4. |emf| =  (area)*|change in B/change in t| = (final flux  -  initial flux)/ change in t. NOW YOU NEED THE AREA, THAT OF A CIRCLE. Note that the final flux is zero since the angle in figure 29.4 is zero. 90. Note: initial flux = B*(area)
5.   Let the field point out of the  page initially.  That means the induced current will be counterclockwise   and the induced magnetic field outward. |emf| =  (area)*(change in B/change in t) + (change in area/change in time)*B, using the calculus rule for products. Note than the (change in  B)/(change in t) refers to the field  magnitude B. Note: (change in B) /(change in time) is given to you and it is negative. Now the (change in area/(change in time) = d(area)/dt = 2*pi*r*dr/dt.  Set | emf | = 0 and solve for the rate of change of r.
 
6.  See #1 but change things a little. The flux changes uniformly. But let the field point out of the page initially and inward finally. That means the induced current will be counterclockwise   and the induced magnetic field outward.
|emf| =  (area)*|change in B/change in t| =
(final flux  -  initial flux)/ change in t. The area is needed.
8. (a) R increases, current I decreases in magnitude, which means that the field magnitude  caused by the current decreases as well.  Now, the above mentioned field points outward from example 28-12. Thus, for the inner loop, the external field points out and decreases in magnitude. Thus the induced field within the inner loop points out to oppose that decrease. What is the direction of the inner loop's current, CW or CCW?
(b) Think about it carefully.
9. Rotate the picture so that the  loop plane lies  in the plane of the page. The South pole of the external solenoid is nearer to the loop; that means the external field  through the loop is  out the page. The magnitude of the filed is decreasing. Thus,  the current in the loop will be counterclockwise  and the induced magnetic field  thus points out of the page. FOR EXTRA UNDERSTANDING RECALL THE FARADAY'S LAB.
10.  See #1 but change things a little. The flux changes uniformly. But let the field point out of the page initially and inward finally. That means the induced current will be counterclockwise   and the induced magnetic field outward.
|emf| =  (area)*|change in B/change in t| =
(final flux  -  initial flux)/ change in t. The area is needed.
11. |emf| = B*d(area)/dt. = B*2*pi*r*dr/dt. , where dr/dt is given  to you.
29. Extremely important problem. First  of all,. you know  the current will be counterclockwise in the circuit .  That's because the flux is increasing as the sliding rod moves right. Inside the sliding rod, the charges move upward under the influence of the magnetic force. Note that inside  the sliding rod,     the net force on the upward moving charge is zero as it drifts upward in a state of dynamic equilibrium established by the rightward external force applied to the rod. The upward magnetic force is balanced by the downward electric force built up by the  collection of positive charges on the top end and the negative  charges  on the bottom end of the sliding rod as well as downward resistance forces. The charges then move through the U shaped conductor under the influence of the electric field established by the charges at either end of the sliding rod; note: inside the U-shaped conductor, the charges are also in dynamic equilibrium with the electric force balanced by resistance forces opposite the constant drift velocity.
(a)  See  example 29-8.
(b ) I emf/(total resistance) ; total resistance includes both the sliding rod and the U-shaped conductor-- note that the  U-shaped conductor 's resistance is increasing since its effective  length is growing. That is why the U-shaped conductor's resistance  is considered to be instantaneous for a particular rod position.
(c) External force = ILB, where L is the length of the sliding rod,  B is the external  magnetic field magnitude, and I is the induced current at the sliding rod's instantaneous location.  This force is required to keep the charges moving in dynamic equilibrium throughout the circuit.
54. The magnetic field points in and is decreasing in magnitude; that means the sense of the emf along the loop is clockwise since the induced field must be in to keep the flux strong. That means the electric field vector is also clockwise in its orientation. That means for example the electric field at Q's location  points down, TANGENT TO THE CIRCLE AT THAT POINT.   Thus, what direction is the force on Q?
DISCUSSIONS TO BE CONTINUED