(d) Estimate the average current through resistor C .5 . A loop of wire is located inside a uniform magnetic field, as shown above. Name at least four things
you could do to induce a current in the loop.
Solutions to Practice Problems
1 . C —Use the right-hand rule for the force on charged particles. You point in the direction of the
velocity, and curl your fingers in the direction of the magnetic field. This should get your thumb
pointing into the page. Because this is a positive charge, no need to switch the direction of the force.
2 . C —Use the right-hand rule for the force on a wire. Look at each part of this wire. At the leftmost
and rightmost points, the current is along the magnetic field lines. Thus, these parts of the wire
experience no force. The topmost part of the wire experiences a force out of the page (point to the
right, fingers curl up the page, the thumb points out of the page). The bottommost part of the wire
experiences a force into the page. So, the wire will rotate.
3 . A —Use the right-hand rule for the force on a charge. Point in the direction of velocity, curl the
fingers into the page, the thumb points up the page ... but this is a negative charge, so the force on the
charge is down the page. Now, the electric force must cancel the magnetic force for the charge to
move in a straight line, so the electric force should be up the page. (E and B fields cannot cancel, but
forces sure can.) The direction of an electric force on a negative charge is opposite the field; so the
field should point down, toward the bottom of the page.
4 . (a) Flux equals zero because the field points along the loop, not ever going straight through the loop.
(b) Flux is maximum when the field is pointing straight through the loop; that is, when the loop is
perpendicular to the page. Then flux will be just BA = 5.0 × 10−5 T·π(0.20 m)^2 = 6.3 × 10−6 T·m^2 .
(Be sure your units are right!)
(c) Induced EMF for this one loop is change in flux over time interval. It takes 1/500 of a second for
the loop to make one complete rotation; so it takes ^1 / 4 of that, or 1/2000 of a second, for the loop
to go from zero to maximum flux. Divide this change in flux by 1/2000 of a second ... this is 6.3 ×
10 −6 T·m^2 /0.0005 s = 0.013 V. (That’s 13 mV.)
(d) Now we can treat the circuit as if it were attached to a battery of voltage 13 mV. The equivalent
resistance of the parallel combination of resistors B and C is 5 Ω; the total resistance of the circuit
is 15 Ω. So the current in the whole circuit is 0.013 V/15 W = 8.4 × 10−4 A. (This can also be
stated as 840 μA.) The current splits evenly between resistors B and C since they’re equal
resistances, so we get 420 μA for resistor C .