Hints and answers to evens:

• Chapter 28: Question 6: Hint: The magnetic force is perpendicular to the particle's velocity at all times. What does a force perpendicular to the path of a moving mass do to it? Recall some examples from Physics 174.
  Answer:It curves the particle's path without changing its speed. This is similar to the normal force of the road on a car, or of tension in a string on a tetherball, or any other force perpendicular to the motion.
• Chapter 28: Exercises and Problems:
  • 5: Try decomposing each of the velocity vectors in parts a-e into components; for example the velocity at point d is (x-hat - z-hat / sqrt(2)). Then take the cross product with B = B x-hat.
    Recall that x-hat × x-hat = 0 = y-hat × y-hat = z-hat × z-hat ; and:
    x-hat × y-hat = z-hat
    y-hat × x-hat = - z-hat, and so on.
  • 11: Something like this MIGHT be on the test:
    Calculate the total flux coming OUT of the prism shown; the vertical sides should give a very very simple answer and then you need merely calculate the flux out through each triangle and add them. Compare the result to "Gauss' Law for Magnetism" to decide if the field given is physically plausible.
  • 13: Something like this WILL be on the test. See also Examples 28-3 and 28-4:
    This should be easy; recall that you can set the magnitude of the centripetal force on the electron equal to the magnitude of the magnetic force on it:
    m v²/R = F_C = q v_{perpendicular}B
    yielding: m vq B = R. Now Solve for R.
    To find the direction, note that it must be directly into or out of the page (why?) and then use the right hand rule to decide which by figuring the force on the electron in either case. Remember the force must point toward the center of the circle. (There is a short cut - the "left hand rule" but do not use this unless you really understand what is going on, otherwise you will just confuse yourself.)
  • 23: Something like this MIGHT be on the test:
    This is very similar to the preceding problem, only you are solving for mass not magnetic field. If this is the test problem, I'd show the path and the B-field and ask you to determine either the mass or the speed and also to determine the sign of the charge from the direction the path curves. The only tricky part in this problem is that they give the DIAMETER of the circle, NOT the radius.
  • 25: This exact problem will not be on the test, but calculating a force on a piece of straight wire will be a problem element, so learn how to do it!
    Hint: the integral of I dl × B along a straight piece of wire is just I L B sin q, where q is the angle between the current and the magnetic field, measured counterclockwise around the force vector.
  • 32: Something like this but simpler WILL be on the test.
    For parts (a) and (c), the z-axis sticks OUT of the page; for parts (b) and (d) the y axis sticks INTO the page. If I'd realized they hadn't drawn these axes on the diagrams, I would have not assigned this problem. You may add these axes in your books, it's not defacing them, it's improving them! With that said, now you want to get the magnetic moment of each coil. Put your left hand in your pocket. Put your right hands fingers curling in the direction of the current in the coil. Your right thumb will stick in the direction of the coil's magnetic moment m. The magnitude of the magnetic moment equals N I A. Now recall the relations between magnetic moment and torque and do some cross products; recall that between magnetic moment and energy. These are given in the text and in the chapter summary.
    Answer:(a): torque = -NIAB x-hat, U=0. (b)torque = 0, U=-NIAB. (c)torque = +NIAB x-hat, U=0. (d)torque = 0, U=+NIAB.
  • 42: This exact problem will not be on the test but elements of it may be. (i.e. you must be able to calculate accelerations of a particle due to E or B fields, and apply simple kinematics).
    If you know how to do a Taylor series expansion of sine and cosine functions, you can just do that to the real circular path of the particle. However, if you DON'T know how to do that (this is not a math class) you can solve this problem by the "trick" of using the fact path is NEARLY straight and the magnetic force is NEARLY constant. Dividing the problem into steps:
    • Find the speed with which the particle leaves the electron gun. We did this as a computer exercise; it is isomorphic to a roller coaster problem.
    • Find the initial magnetic force on the particle as it enters the region of the magnetic field.
    • Pretend that the initial magnetic force on the particle remains constant and this now becomes a constant acceleration problem, identical to a projectile motion problem from Physics 174. Solve.
    
    
  • 42. This was shown in class - either solve for the radius of curvature of the electron's path R = mv/qB = m * sqrt (2*e*V/m)/e*B = sqrt (2*m*V/e)/B, then draw an arc of a circle of that radius which subtends an angle of arctan(D/R) and solve for the drop d with a Taylor series approximation, or treat it as a constant acceleration problem where the acceleration = e*v*B/m=sqrt(2*e*V/m)*B/m and the distance traversed is D.
  • 63: Something like this MIGHT be on the test. In any case some form of rectangular or square current loop WILL be on the test. If the problem is this hard it would be among the hardest on the test (mathematically).
    a. I'll try to show a picture of what this might look like.
    b. Calculate the force per unit length, then integrate THAT. This should not be two hard - it's a linear function. Be careful about components.
    c. Just add up the contributions for each side - vector sum.
  • Chapter 29: Questions:
    • 4: Think about the details of electric current- it is driven by an electric field. A mechanical analogy might be useful: consider a gravitational field driving particles down a curved track. The normal force of the track does no work (like a magnetic field) but converts vertical motion (driven by the work-doing field) into horizontal motion.
    • 9: You can think in terms of fields and currents or just in terms of parallel and antiparallel currents.
    • Chapter 29: Exercises:
      
      Exercises 3, 13, 14
      • 3: Use the Biot-Savart law; look it up if needed.
      • 13: Use B = mu₀ I / 2 pi R around each wire and add up the fields.
      • 14: Use B = mu₀ I / 2 pi R around each wire and add up the fields produced by each wire. Then use force = current times vector dL cross B...