27.43. Identify: The period is , the current is and the magnetic moment is

Set Up: The electron has charge . The area enclosed by the orbit is .

Execute: (a)

(b) Charge passes a point on the orbit once during each period, so .

(c)

Evaluate: Since the electron has negative charge, the direction of the current is opposite to the direction of motion of the electron.

27.45. Identify: The magnetic field exerts a torque on the current-carrying coil, which causes it to turn. We can use the rotational form of Newton’s second law to find the angular acceleration of the coil.

Set Up: The magnetic torque is given by , and the rotational form of Newton’s second law is . The magnetic field is parallel to the plane of the loop.

Execute: (a) The coil rotates about axis A₂ because the only torque is along top and bottom sides of the coil.

(b) To find the moment of inertia of the coil, treat the two 1.00-m segments as point-masses (since all the points in them are 0.250 m from the rotation axis) and the two 0.500-m segments as thin uniform bars rotated about their centers. Since the coil is uniform, the mass of each segment is proportional to its fraction of the total perimeter of the coil. Each 1.00-m segment is 1/3 of the total perimeter, so its mass is (1/3)(210 g) = 70 g = 0.070 kg. The mass of each 0.500-m segment is half this amount, or 0.035 kg. The result is

The torque is

Using the above values, the rotational form of Newton’s second law gives

Evaluate: This angular acceleration will not continue because the torque changes as the coil turns.

28.25. Identify: Apply Eq.(28.11).

Set Up: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other.

Execute: (a) and the force is repulsive since the currents are in opposite directions.

(b) Doubling the currents makes the force increase by a factor of four to

Evaluate: Doubling the current in a wire doubles the magnetic field of that wire. For fixed magnetic field, doubling the current in a wire doubles the force that the magnetic field exerts on the wire.

28.28. Identify: Apply Eq.(28.11) for the force from each wire.

Set Up: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other.

Execute: On the top wire upward. On the middle wire, the magnetic forces cancel so the net force is zero. On the bottom wire downward.

Evaluate: The net force on the middle wire is zero because at the location of the middle wire the net magnetic field due to the other two wires is zero.

28.31. Identify: Calculate the magnetic field vector produced by each wire and add these fields to get the total field.

Set Up: First consider the field at P produced by the current in the upper semicircle of wire. See Figure 28.31a.

	Consider the three parts of this wire a: long straight section, b: semicircle c: long, straight section
Figure 28.31a

Apply the Biot-Savart law to each piece.

Execute: part a See Figure 28.31b.

	so
Figure 28.31b

The same is true for all the infinitesimal segments that make up this piece of the wire, so B = 0 for this piece.

part c See Figure 28.31c.

	so for this piece.
Figure 28.31c

part b See Figure 28.31d.

	is directed into the paper for all infinitesimal segments that make up this semicircular piece, so is directed into the paper and (the vector sum of the is obtained by adding their magnitudes since they are in the same direction).
Figure 28.31d

The angle between the radius of the semicircle. Thus

(We used that is equal to the length of wire in the semicircle.) We have shown that the two straight sections make zero contribution to so and is directed into the page.

	For current in the direction shown in Figure 28.31e, a similar analysis gives out of the paper
Figure 28.31e

are in opposite directions, so the magnitude of the net field at P is

Evaluate: When

28.33. Identify: Apply Eq.(28.16).

Set Up: At the center of the coil, . a is the radius of the coil, 0.020 m.

Execute: (a)

(b)

Evaluate: As shown in Figure 28.41 in the textbook, the field has its largest magnitude at the center of the coil and decreases with distance along the axis from the center.

28.35. Identify: Apply Ampere’s law.

Set Up: 

Execute: (a) and

(b) since at each point on the curve the direction of is reversed.

Evaluate: The line integral around a closed path is proportional to the net current that is enclosed by the path.

28.43. Identify and Set Up: Use the appropriate expression for the magnetic field produced by each current configuration.

Execute: (a) so .

(b) so .

(c) so .

Evaluate: Much less current is needed for the solenoid, because of its large number of turns per unit length.