27.15.     (a) Identify: Apply Eq.(27.2) to relate the magnetic force  to the directions of  The electron has negative charge so  is opposite to the direction of  For motion in an arc of a circle the acceleration is toward the center of the arc so  must be in this direction.

Set Up: 

As the electron moves in the semicircle, its velocity is tangent to the circular path. The direction of  at a point along the path is shown in Figure 27.15.

Figure 27.15

 

Execute: For circular motion the acceleration of the electron  is directed in toward the center of the circle. Thus the force  exerted by the magnetic field, since it is the only force on the electron, must be radially inward. Since q is negative,  is opposite to the direction given by the right-hand rule for  Thus  is directed into the page. Apply Newton's 2nd law to calculate the magnitude of  

(b) Identify and Set Up: The speed of the electron as it moves along the path is constant. ( changes the direction of  but not its magnitude.) The time is given by the distance divided by

Execute: The distance along the semicircular path is  so

Evaluate: The magnetic field required increases when v increases or R decreases and also depends on the mass to charge ratio of the particle.

         

27.17.     Identify and Set Up: Use conservation of energy to find the speed of the ball when it reaches the bottom of the shaft. The right-hand rule gives the direction of  and Eq.(27.1) gives its magnitude. The number of excess electrons determines the charge of the ball.

Execute: 

speed at bottom of shaft:

 is downward and  is west, so  is north. Since  is south.

Evaluate: Both the charge and speed of the ball are relatively small so the magnetic force is small, much less than the gravity force of 1.5 N.

 

27.33.     Identify: The magnetic force is  For the wire to be completely supported by the field requires that and that and are in opposite directions.

Set Up: The magnetic force is maximum when  The gravity force is downward.

Execute: (a)   This is a very large current and ohmic heating due to the resistance of the wire would be severe; such a current isn’t feasible.

(b) The magnetic force must be upward. The directions of I,  and  are shown in Figure 27.33, where we have assumed that is south to north. To produce an upward magnetic force, the current must be to the east. The wire must be horizontal and perpendicular to the earth’s magnetic field.

Evaluate: The magnetic force is perpendicular to both the direction of I and the direction of

Figure 27.33

27.37.     Identify: .

Set Up: Since the field is perpendicular to the rod it is perpendicular to the current and .

Execute: 

Evaluate: The force and current are proportional. We have assumed that the entire 0.200 m length of the rod is in the magnetic field.

 

28.1.      Identify and Set Up: Use Eq.(28.2) to calculate  at each point.

 is the vector from the charge to the point where the field is calculated.

Execute: (a)

(b)

(c)

(d)

Evaluate: At each point  is perpendicular to both  and  B = 0 along the direction of

28.3.      Identify: A moving charge creates a magnetic field.

Set Up: The magnetic field due to a moving charge is .

Execute: Substituting numbers into the above equation gives

(a)

B = 6.00 ´ 10–8 T, out of the paper, and it is the same at point B.

(b) B = (1.00 ´ 10–7 Tm/A)(1.60 ´ 10–19 C)(3.00 ´ 107 m/s)/(2.00 ´ 10–6 m)2

B = 1.20 ´ 10–7 T, out of the page.

(c) B = 0 T since sin(180°) = 0.

Evaluate: Even at high speeds, these charges produce magnetic fields much less than the Earth’s magnetic field.

         

28.9.      Identify: A current segment creates a magnetic field.

Set Up: The law of Biot and Savart gives .

Execute: Applying the law of Biot and Savart gives

(a)  = 4.40 ´ 10–7 T, out of the paper.

(b) The same as above, except  and f = arctan(5/14) = 19.65°, giving dB = 1.67 ´ 10–8 T, out of the page.

(c) dB = 0 since f = 0°.

Evaluate: This is a very small field, but it comes from a very small segment of current.

 

28.12.     Identify: A current segment creates a magnetic field.

Set Up: The law of Biot and Savart gives .

Both fields are into the page, so their magnitudes add.

Execute: Applying the law of Biot and Savart for the 12.0-A current gives

 = 8.79 ´ 10–8 T

The field from the 24.0-A segment is twice this value, so the total field is 2.64 ´ 10–7 T, into the page.

Evaluate: The rest of each wire also produces field at P. We have calculated just the field from the two segments that are indicated in the problem.

         

28.15.     Identify: We can model the lightning bolt and the household current as very long current-carrying wires.

Set Up: The magnetic field produced by a long wire is .

Execute: Substituting the numerical values gives

(a) B =  = 8 ´ 10–4 T

(b)= 4.0 ´ 10–5 T.

Evaluate: The field from the lightning bolt is about 20 times as strong as the field from the household current.

         

28.21.     Identify: . The direction of  is given by the right-hand rule in Section 20.7.

Set Up: Call the wires a and b, as indicated in Figure 28.21. The magnetic fields of each wire at points P1 and P2 are shown in Figure 28.21a. The fields at point 3 are shown in Figure 28.21b.

Execute: (a) At , and the two fields are in opposite directions, so the net field is zero.

(b) . . and are in the same direction so

has magnitude and is directed toward the top of the page.

(c) In Figure 28.21b,  is perpendicular to  and  is perpendicular to .  and . and .

B has magnitude  and is directed to the left.

Evaluate: At points directly to the left of both wires the net field is directed toward the bottom of the page.

                

Figure 28.21