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Magnetism, magnetic field of bar magnet

  1. Mar 4, 2008 #1
    Can anyone check if I answered these questions correctly? Thanks a bunch!

    1. The N pole of a bar magnet is pointed in along the axis of a horizontal, continuous coil of wire. When the magnet is quickly roated about a vertical axis until the opposite end of the madnet points into the coil, when looking past the magnet at the coils, the induced current in the loops of wire will be:

    a. clockwise
    c.greater than if the magnet is just pulled away
    d.both a & b
    e. both a & c

    2. The rectangle below represents a block of material through which positive charge carriers move. When the left end is connected to the + terminal of a battery and the right end is connected to the - terminal of this battery a current shall flow through the metal.


    If a magnetic field is applied so that it's perpendicular to the plane of this paper and directed down toward it, an induced emf will be stablished with + on the:
    a.top face of the block and - on the bottom face
    b.bottom face of the block and - on the top face
    c.face of the block nearest you and - on the face opposite this
    d.right end face of the block and - on the opposite, left end face
    e.none of these

    3. Electrons move with a drift velocity of 1e-5m/s through a copper wire when a 5 A current flows through it. This wire has a diameter of 3.2 mm/ When the wire is held so it's perpendicular to the earths's B of the 5e-5 T, what will be the magnitude of the induced Hall emf?
    a. zero
    b. 10^-12 V
    c. 10^-10 V
    d. 10^-7 V
    E. none of these
  2. jcsd
  3. Mar 5, 2008 #2


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    If I'm interpreting the phrasing correctly, we are considering the situation where the flux from the north pole of the magnet starts out pointing perpendicularly into the coil (away from the magnet) and is dying away as the magnet is turned. As viewed from the magnet, then, the level of magnetic flux into the coil pointed away from the magnet is decreasing. Lenz' Law (basically a shorthand for dealing with Faraday's Law) tells us that the direction of the induced current will circulate to generate a magnetic field through the coil that resists the change in external flux. Since the flux pointing away from the magnet is decreasing, the induced current will circulate in a direction to "compensate" by producing more flux pointing away from the magnet. The "right-hand rule" for a current loop says that flux pointing away from the magnet is produced by a clockwise circulation of current, as seen from the magnet.

    So I believe the answer is (A). [I can see that it would be easy to get confused by the phrasing of the problem, in the absence of a diagram.]

    The potential from the battery is driving a current flowing from left to right in the conductive bar, as depicted in the drawing. The magnetic field lines are supposed to be perpendicular to the monitor screen we are looking at and pointing away from us. The magnetic field produces a magnetic force given by IL x B , that is, the direction of the magnetic force is given by the cross product of the current direction and the magnetic field direction. The "right-hand rule" for cross products tells us that the magnetic force will be "upward" (toward the top of the screen), which corresponds to the top of the bar in the drawing.

    So the positive charge carriers experience a second force driving them from the bottom to the top of the bar. This is analogous to having an electric potential which is higher at the bottom of the bar than at the top (as if there were an electric field pointing toward the top). [We call this an "induced emf" because it is not a true potential function, so we fall back on the antique term "electromotive force".] I believe the answer will be (B).

    The "drift current" of the electrons experiences a magnetic force from the earth's magnetic field, since it is oriented perpendicularly to it; the magnitude of that force on an electron will be given by evB . The strength of the "field" corresponding to the force would be evB / e = vB , which gives

    vB = (1.0·10^-5 m/sec)(5.0·10^-5 T) = 5.0·10^-10 N/C or V/m .

    The "potential" across the width of the wire, which gives us the "induced emf", would then be

    vB · w = (5.0·10^-10 V/m)(0.0032 m) = 1.6·10^-12 V .

    The closest order-of-magnitude choice is (B). If this is from a standardized test (like the GRE), this is acceptable agreement. (If this were a physics course exam, they probably would have given a couple significant figures of precision in the choices...) If this is not a sufficient level of agreement, then I would have to go with (E) -- none of these.
    Last edited: Mar 5, 2008
  4. Mar 5, 2008 #3
    Thanks. I was going between A & B for #1 but I never got the picture straight in my head. It was a Physics exam so I guess I got #3 correct. And I actually marked A for #2....I thought the velocity is going toward the positive which makes the force downward because the currents goes out of + and into -....does that make any sense?
  5. Mar 5, 2008 #4


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    The drift velocity is to the right because the charge carriers are following the electric field imposed by the battery. The so-called induced or Hall emf is the result of the magnetic force on the charge carriers produced by the external magnetic field. In problem 2, this force is toward the top of the bar and is treated as if there were a second voltage source present. Since the positive charge carriers are deflected toward the top, this "potential" acts as if it were oriented with "positive" at the bottom and "negative" at the top.
    Last edited: Mar 5, 2008
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