29.1. Identify: Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the coil.

Set Up: The flux through a coil of N turns is F = NBAcosf, and by Faraday’s law the magnitude of the induced emf is = dF/dt.

Execute: (a) DF = NBA = (50)(1.20 T)(0.250 m)(0.300 m) = 4.50 Wb

(b) = dF/dt = (4.50 Wb)/(0.222 s) = 20.3 V

Evaluate: This induced potential is certainly large enough to be easily detectable.

29.5. Identify: Apply Faraday’s law.

Set Up: Let +z be the positive direction for. Therefore, the initial flux is positive and the final flux is zero.

Execute: (a) and (b) Since is positive and is toward us, the induced current is counterclockwise.

Evaluate: The shorter the removal time, the larger the average induced emf.

29.6. Identify: Apply Eq.(29.4).

Set Up: 

Execute: (a)

(b) At

Evaluate: The rate of change of the flux is increasing in time, so the induced current is not constant but rather increases in time.

29.7. Identify: Calculate the flux through the loop and apply Faraday’s law.

Set Up: To find the total flux integrate over the width of the loop. The magnetic field of a long straight wire, at distance r from the wire, is . The direction of is given by the right-hand rule.

Execute: (a) When , into the page.

(b)

(c)

(d)

(e)

Evaluate: The induced emf is proportional to the rate at which the current in the long straight wire is changing

29.11. Identify: A change in magnetic flux through a coil induces an emf in the coil.

Set Up: The flux through a coil is F = NBAcosf and the induced emf is = dF/dt.

Execute: (a) = dF/dt = d[A(B₀ + bx)]/dt = bA dx/dt = bAv

(b) clockwise

(c) Same answers except the current is counterclockwise.

Evaluate: Even though the coil remains within the magnetic field, the flux through it increases because the strength of the field is increasing.

29.15. Identify and Set Up: The field of the induced current is directed to oppose the change in flux.

Execute: (a) The field is into the page and is increasing so the flux is increasing. The field of the induced current is out of the page. To produce field out of the page the induced current is counterclockwise.

(b) The field is into the page and is decreasing so the flux is decreasing. The field of the induced current is into the page. To produce field into the page the induced current is clockwise.

(c) The field is constant so the flux is constant and there is no induced emf and no induced current.

Evaluate: The direction of the induced current depends on the direction of the external magnetic field and whether the flux due to this field is increasing or decreasing.

29.17. Identify and Set Up: Apply Lenz's law, in the form that states that the flux of the induced current tends to oppose the change in flux.

Execute: (a) With the switch closed the magnetic field of coil A is to the right at the location of coil B. When the switch is opened the magnetic field of coil A goes away. Hence by Lenz's law the field of the current induced in coil B is to the right, to oppose the decrease in the flux in this direction. To produce magnetic field that is to the right the current in the circuit with coil B must flow through the resistor in the direction a to b.

(b) With the switch closed the magnetic field of coil A is to the right at the location of coil B. This field is stronger at points closer to coil A so when coil B is brought closer the flux through coil B increases. By Lenz's law the field of the induced current in coil B is to the left, to oppose the increase in flux to the right. To produce magnetic field that is to the left the current in the circuit with coil B must flow through the resistor in the direction b to a.

(c) With the switch closed the magnetic field of coil A is to the right at the location of coil B. The current in the circuit that includes coil A increases when R is decreased and the magnetic field of coil A increases when the current through the coil increases. By Lenz's law the field of the induced current in coil B is to the left, to oppose the increase in flux to the right. To produce magnetic field that is to the left the current in the circuit with coil B must flow through the resistor in the direction b to a.

Evaluate: In parts (b) and (c) the change in the circuit causes the flux through circuit B to increase and in part (a) it causes the flux to decrease. Therefore, the direction of the induced current is the same in parts (b) and (c) and opposite in part (a).

29.21. Identify: A conductor moving in a magnetic field may have a potential difference induced across it, depending on how it is moving.

Set Up: The induced emf is = vBLsinf, where f is the angle between the velocity and the magnetic field.

Execute: (a) = vBLsinf = (5.00 m/s)(0.450 T)(0.300 m)(sin90°) = 0.675 V

(b) The positive charges are moved to end b, so b is at the higher potential.

(c) E = V/L = (0.675 V)/(0.300 m) = 2.25 V/m. The direction of is from, b to a.

(d) The positive charge are pushed to b, so b has an excess of positive charge.

(e) (i) If the rod has no appreciable thickness, L = 0, so the emf is zero. (ii) The emf is zero because no magnetic force acts on the charges in the rod since it moves parallel to the magnetic field.

Evaluate: The motional emf is large enough to have noticeable effects in some cases.

29.25. Identify and Set Up:  Use Lenz's law to determine the direction of the induced current. The force required to maintain constant speed is equal and opposite to the force that the magnetic field exerts on the rod because of the current in the rod.

Execute: (a)

(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 induced current must be counterclockwise, so from b to a in the rod.

(c) . is to the left. To keep the bar moving to the right at constant speed an external force with magnitude and directed to the right must be applied to the bar.

(d) The rate at which work is done by the force is The rate at which thermal energy is developed in the circuit is These two rates are equal, as is required by conservation of energy.

Evaluate: 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.