Problem#1
The conducting rod ab shown in Fig. 1 makes contact with metal rails ca and db. The apparatus is in a uniform magnetic field of 0.800 T, perpendicular to the plane of the figure (a) Find the magnitude of the emf induced in the rod when it is moving toward the right with a speed (b) In what direction does the current flow in the rod? (c) If the resistance of the circuit abdc is (assumed to be constant), find the force (magnitude and direction) required to keep the rod moving to the right with a constant speed of You can ignore friction. (d) Compare the rate at which mechanical work is done by the force with the rate at which thermal energy is developed in the circuit (I2R). |
| Fig.1 |
Answer:
Use Lenz’s law to determine the direction of the induced current. The force Fext requaired to maintain constant speed is equal and opposite to the force FI that the magnetic field exert on the rod because of the current in the rod.
(a) the magnitude of the emf induced in the rod when it is moving toward the right with a speed
ε = Vbl = (7.50 m/s)(0.800 T)(0.500 m) = 3.00 V
(b) B is into the page. The flux increases as the bar moves to the right, so the magnetic field of the induced current is out of the page inside the circuit. To produce magnetic field in this direction the current must be counterclockwise, so from b to a in the rod.
(c) the magnitude induced current is
I = ε/R = 3.00 V/1.5 Ω = 2.00 A. Then,
FI = ILB sinφ = (2.00 A)(0.500 m)(0.800 T)sin 90° = 0.800 N
FI is to the left. To keep the bar moving to the right at constant speed an external force with magnitude Fext = 0.800 N and directed to the right must be applied to the bar.
(d) The rate at which work is done by the force Fext is P = Fextv = (0.800 N)(7.50 m/s) = 6.00W
The rate at which thermal energy is developed in the circuit is P = I2R = (2.00 A)2(1.50 Ω) = 6.00W.
These two rates are equal, as is required by conservation of energy.
The force on the rod due to the induced current is directed to oppose the motion of the rod. This agrees with Lenz’s law.
Problem#2
A 1.50-m-long metal bar is pulled to the right at a steady 5.0 m/s perpendicular to a uniform, 0.750-T magnetic field. The bar rides on parallel metal rails connected through a 25.0 Ω resistor, as shown in Fig. 2, so the apparatus makes a complete circuit. You can ignore the resistance of the bar and the rails. (a) Calculate the magnitude of the emf induced in the circuit. (b) Find the direction of the current induced in the circuit (i) using the magnetic force on the charges in the moving bar; (ii) using Faraday’s law; (iii) using Lenz’s law. (c) Calculate the current through the resistor.
 |
| Fig.2 |
Answer:
The moving bar has a motion emf induced across its ends, so it causes a current to flow.
The induced potential is ε = vBL and Ohm’s law is ε = IR.
(a) the magnitude of the emf induced in the circuit is
ε = vBL = (5.0 m/s)(0.750 T)(2.50 m) = 5.6 V
(b) (i) Let q be a positive charge in the moving bar, as shown in Figure 3a. The magnetic force on this charge is F = qv × B which points upward. This force pushes the current in a counterclockwise direction through the circuit.
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| Fig.3 |
(ii) ΦB is positive and is increasing in magnitude, so dΦB/dt > 0. Then by Faraday’s law ε < 0 and the emf and induced current are counterclockwise.
(iii) The flux through the circuit is increasing, so the induced current must cause a magnetic field out of the paper to oppose this increase. Hence this current must flow in a counterclockwise sense, as shown in Figure 8b.
Problem#3
A 0.360-m-long metal bar is pulled to the left by an applied force F. The bar rides on parallel metal rails connected through a 45.0 Ω resistor, as shown in Fig. 3, so the apparatus makes a complete circuit. You can ignore the resistance of the bar and rails. The circuit is in a uniform 0.650-T magnetic field that is directed out of the plane of the figure. At the instant when the bar is moving to the left at 5.90 m/s, (a) is the induced current in the circuit clockwise or counterclockwise and (b) what is the rate at which the applied force is doing work on the bar?
 |
| Fig.3 |
Answer:
The motion of the bar due to the applied force causes a motional emf to be induced across the ends of the bar, which induces a current through the bar. The magnetic field exerts a force on the bar due to this current.
(a) B out of page and ΦB decreasing, so the field of the induced current is out of the page inside the loop and the induced current is counterclockwise.
(b) Combining Fapplied = FB = ILB and ε = BvL,
we have I = ε/R = BvL/R
then, Fapplied = vB2L2/R
The rate at which this force does work is
Papplied = Fappliedv = (vBL)2/R
Papplied = (5.90 m/s x 0.650 T x 0.360 m)2/45.0 Ω = 0.0424 Watt
Problem#4
Consider the circuit shown in Fig. 3, but with the bar moving to the right with speed . As in Problem#3, the bar has length 0.360 m, R = 45.0 Ω, and B = 0.650 T. (a) Is the induced current in the circuit clockwise or counterclockwise? (b) At an instant when the 45.0 Ω resistor is dissipating electrical energy at a rate of 0.840 J/s, what is the speed of the bar?
Answer:
The motion of the bar due to the applied force causes a motional emf to be induced across the ends of the bar, which induces a current through the bar and through the resistor. This current dissipates energy in the resistor.
(a) B is out of the page and ΦB is increasing, so the field of the induced current is into the page inside the loop and the induced current is clockwise.
(b) dissipating electrical energy is
PR = I2R
0.840 J/s = I2 x 45.0 Ω
I = 0.1366 A
So that, the speed of the bar is
I = ε/R = BvL/R
v = IR/BL = (0.1366 A)(45.0 Ω)/(0.650 T x 0.360 m) = 26.3 m/s
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