SOURCES OF MAGNETIC FIELD 28.12. IDENTIFY: A current segment creates a magnetic field. SET UP: The law of Biot and Savart gives 0 2sin4 IdldB r . 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 7 2 2 . 5 0 c m( 1 2 . 0 A ) ( 0 . 0 0 1 5 0 m )4 π 1 0 T m /A 8 . 0 0 c m 4 π ( 0 .0 8 0 0 m )dB = 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. 28.22. IDENTIFY: Use Eq.(28.9) and the right-hand rule to determine points where the fields of the two wires cancel. (a) SET UP: The only place where the magnetic fields of the two wires are in opposite directions is between the wires, in the plane of the wires. Consider a point a distance x from the wire carrying 2 tot75.0 A. IB will be zero where 12.BB EXECUTE: 0 1 0 22 (0.400 m ) 2IIxx 2 1 1 2( 0 . 4 0 0 m ) ; 2 5 . 0 A , 7 5 . 0 AI x I x I I tot0.300 m; 0xB along a line 0.300 m from the wire carrying 75.0 A and 0.100 m from the wire carrying current 25.0 A. (b) SET UP: Let the wire with 1 25.0 AI be 0.400 m above the wire with 2 75.0 A.I The magnetic fields of the two wires are in opposite directions in the plane of the wires and at points above both wires or below both wires. But to have 12BB must be closer to wire #1 since 12,II so can have tot 0B only at points above both wires. Consider a point a distance x from the wire carrying 1 tot25.0 A. IB will be zero where 12.BB EXECUTE: 0 1 0 22 2 (0.400 m )IIxx 21 ( 0 .4 0 0 m ) ; 0 .2 0 0 mI x I x x tot 0B along a line 0.200 m from the wire carrying current 25.0 A and 0.600 m from the wire carrying current 2 75.0 A.I EVALUATE: For parts (a) and (b) the locations of zero field are in different regions. In each case the points of zero field are closer to the wire that has the smaller current. 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. 28 EXECUTE: On the top wire 22 0011 ,2 2 4F I IL d d d upward. On the middle wire, the magnetic forces cancel so the net force is zero. On the bottom wire 22 0011 ,2 2 4F I IL d d d 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. 28.30. IDENTIFY: The magnetic field at the center of a circular loop is 02IB R . By symmetry each segment of the loop that has length l contributes equally to the field, so the field at the center of a semicircle is 12 that of a full loop. SET UP: Since the straight sections produce no field at P, the field at P is 04IB R . EXECUTE: 04IB R . The direction of B is given by the right-hand rule: B is directed into the page. EVALUATE: For a quarter-circle section of wire the magnetic field at its center of curvature is 08IB R . 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 1I 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 00 23ˆ44Id Idd rr l r l rB = = to each piece. EXECUTE: part a See Figure 28.31b. 0,dl r= so 0dB 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. 0,dl r= so 0 and 0dB B for this piece. Figure 28.31c part b See Figure 28.31d. dlr is directed into the paper for all infinitesimal segments that make up this semicircular piece, so B is directed into the paper and B dB (the vector sum of the dB is obtained by adding their magnitudes since they are in the same direction). Figure 28.31d sin .d rdl lr The angle between and is 90 and ,d r Rlr the radius of the semicircle. Thus d Rdllr 0 0 1 0 13 3 24 4 4Id I R Id B d l d lr R R lr 0 1 0 1 0 122 ()4 4 4I I IB d B d l RR R R (We used that dl is equal to ,R the length of wire in the semicircle.) We have shown that the two straight sections make zero contribution to ,B so 1 0 1/4B I R and is directed into the page. For current in the direction shown in Figure 28.31e, a similar analysis gives 2 0 2 /4 ,B I R out of the paper Figure 28.31e 12 and BB are in opposite directions, so the magnitude of the net field at P is 0 1 212 .4IIB B B R EVALUATE: When 12, 0.I I B 28.46. IDENTIFY: Use Eq.(28.24), with 0 replaced by m0K , with m 80.K SET UP: The contribution from atomic currents is the difference between B calculated with and B calculated with 0. EXECUTE: (a) m 0 0 ( 8 0 ) ( 4 0 0 ) (0 . 2 5 A ) 0 . 0 2 6 7 T .2 2 2 (0 . 0 6 0 m )N I K N IB rr (b) The amount due to atomic currents is 7 9 7 9 (0 .0 2 6 7 T ) 0 .0 2 6 3 T .8 0 8 0BB EVALUATE: The presence of the core greatly enhances the magnetic field produced by the solenoid. 28.59. IDENTIFY: Use Eq.(28.9) and the right-hand rule to calculate the magnitude and direction of the magnetic field at P produced by each wire. Add these two field vectors to find the net field. (a) SET UP: The directions of the fields at point P due to the two wires are sketched in Figure 28.59a. EXECUTE: 12 and BB must be equal and opposite for the resultant field at P to be zero. 2B is to the right so 2I is out of the page. Figure 28.59a 0 1 0 0 2 0 212 12 6 . 0 0 A2 2 1 . 5 0 m 2 2 0 . 5 0 mI I IBBrr 0 0 212 6 . 0 0 A s a y s 2 1 . 5 0 m 2 0 . 5 0 mIBB 2 0 .5 0 m 6 .0 0 A 2 .0 0 A1 .5 0 mI (b) SET UP: The directions of the fields at point Q are sketched in Figure 28.59b. EXECUTE: 011 12 IB r 761 6 . 0 0 A( 2 1 0 T m / A ) 2 . 4 0 1 0 T0 . 5 0 mB 022 22 IB r 772 2 . 0 0 A( 2 1 0 T m / A ) 2 . 6 7 1 0 T1 . 5 0 mB Figure 28.59b 12 and BB are in opposite directions and 12BB so 6 7 612 2 . 4 0 1 0 T 2 . 6 7 1 0 T 2 . 1 3 1 0 T , a n d B B B B is to the right. (c) SET UP: The directions of the fields at point S are sketched in Figure 28.59c. EXECUTE: 011 12 IB r 761 6 . 0 0 A( 2 1 0 T m / A ) 2 . 0 0 1 0 T0 . 6 0 mB 022 22 IB r 772 2 . 0 0 A( 2 1 0 T m / A ) 5 . 0 0 1 0 T0 . 8 0 mB Figure 28.59c 12 and BB are right angles to each other, so the magnitude of their resultant is given by 2 2 6 2 7 2 612 ( 2 . 0 0 1 0 T ) ( 5 . 0 0 1 0 T ) 2 . 0 6 1 0 TB B B EVALUATE: The magnetic field lines for a long, straight wire are concentric circles with the wire at the center. The magnetic field at each point is tangent to the field line, so B is perpendicular to the line from the wire to the point where the field is calculated. 28.67. IDENTIFY: Find the vector sum of the fields due to each loop. SET UP: For a single loop 2 02 2 3/2 .2( )IaB xa Here we have two loops, each of N turns, and measuring the field along the x-axis from between them means that the “x” in the formula is different for each case: EXECUTE: Left coil: 2 0l 2 2 3 2 .2 2 ( ( 2 ) )a μ N Iax x B x a a Right coil: 2 0r 2 2 3 2 .2 2 ( ( 2 ) )a μ N Iax x B x a a So, the total field at a point a distance x from the point between them is 20 2 2 3 2 2 2 3 211 .2 ( ( 2 ) ) ( ( 2 ) )μ N I aB x a a x a a (b) B versus x is graphed in Figure 28.67. Figure 28.67a is the total field and Figure 27.67b is the field from the right-hand coil. (c) At point P, 0x and 3222 0 0 02 2 3 2 2 2 3 2 2 3 21 1 42 ( ( 2 ) ) ( ( 2 ) ) ( 5 4 ) 5μ N I a μ N I a μ N IB a a a a a a (d) 3 2 3 2 004 4 ( 3 0 0 ) ( 6 . 0 0 A ) 0 . 0 2 0 2 T .5 5 ( 0 . 0 8 0 m )μ N I μB a (e) 2 0 2 2 5 2 2 2 5 23 ( 2 ) 3 ( 2 )2 ( ( 2 ) ) ( ( 2 ) )dB μ N I a x a x ad x x a a x a a . At 0x , 20 2 2 5 2 2 2 5 20 3 ( 2 ) 3 ( 2 ) 02 ( ( 2 ) ) ( ( 2 ) )xdB μ N I a a ad x a a a a . 2 2 2 20 2 2 2 5 2 2 2 7 2 2 2 5 / 2 2 2 7 / 23 6 ( 2 ) ( 5 2 ) 3 6 ( 2 ) ( 5 2 )2 ( ( 2 ) ) ( ( 2 ) ) ( ( 2 ) ) ( ( 2 ) )dB μ N I a x a x ad x x a a x a a x a a x a a At 0x , 5 / 2 2 2 2 20 2 2 2 7 / 2 2 2 5 / 2 2 2 7 / 2220 3 6 ( 2 ) ( 5 2 ) 3 6 ( 2 ) ( 5 2 ) 0.2 ( ( 2 ) ) ( ( 2 ) ) ( ( 2 ) )( ( 2 ) )xdB μ N I a a ad x a a a a a aaa EVALUATE: Since both first and second derivatives are zero, the field can only be changing very slowly. Figure 28.67 28.69. IDENTIFY: Apply 0 2 ˆ4μ Idd πrlrB= . SET UP: The contribution from the straight segments is zero since 0.d lr The magnetic field from the curved wire is just one quarter of a full loop. EXECUTE: 0014 28μ I μ IB RR and is directed out of the page. EVALUATE: It is very simple to calculate B at point P but it would be much more difficult to calculate B at other points. 28.87. IDENTIFY: The rotating disk produces concentric rings of current. Calculate the field due to each ring and integrate over the surface of the disk to find the total field. SET UP: At the center of a circular ring carrying current I, 0 2IB r . EXECUTE: The charge on a ring of radius r is 222.Q rd rq A rd r a If the disk rotates at n turns per second, then the current from that ring is 22dq Q nrdrdI ndqdt a . Therefore, 0 0 0 22222I Q n r d r n Q d rdB r r a a . We integrate out from the center to the edge of the disk and find 00 200 .aa n Q d r n QB d B aa Sally