1.
A bar of length 10 cm slides along metal rails at a speed of 5 m/s in a
magnetic field of 0.1 T. What is the motional emf induced in the bar and
rails?
2. The rectangular loop shown in Fig. 21-44 is pushed into the magnetic field
which points inward as shown. In what direction is the induced current
flowing through the resistor R?
3. The moving rod in Fig. 21-9 is 12.0 cm long and moves with a speed of 15.0
cm/s. If the magnetic field is 0.800 T, calculate (a) the emf developed, and
(b) the electric field in the rod.
4.
4. In Fig. 21-9, the rod moves with a speed of 1.9 m/s, is 30.0 cm long, and has a
resistance of 2.5 Ω. The magnetic field is 0.75 T, and the resistance of the Ushaped conductor is 25.0 Ω at a given instant. Calculate (a) the induced emf, (b)
the current flowing in the circuit, and (c) the external force necessary to ensure
that the rod is moving at constant velocity at that instant.
Once again, let’s just turn 30 cm into .3 m to start with.
a) We’ll use the equation ε=Blv and just plug in all the numbers
ε = Blv = .75 * .3 * 1.9 = .4275 = .43 V
b) Great, current flowing in the circuit. Well, let’s use I=ε/R
I = .4275 / (25 + 2.5) = .01554 = .016 A
c) F = IlB = .01554 * .3 * .75 = .003497 = .0035
5. A single rectangular loop of wire with dimensions shown in Fig. 21-49 is situated
so that part is inside a region of uniform magnetic field of 0.450 T and part is
outside the field. The total resistance of the loop is 0.230 Ω. Calculate the force
required to pull the loop from the field (to the right) at a constant velocity of 3.40
m/s. Neglect gravity.
For this, we’ll start with the equation E=Bvl = .45 * 3.4 * .35 = .5355
V=IR or .5355 = I * .23, so I = .5355/.23 = 2.328
And for the grand finale, F = IlB = 2.328 * .35 * .45 = .366666 = .367 N