hw7 - Problem 2. A sing le-turn w ire loop is 2.0 cm in d...

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Problem 2. A single-turn wire loop is 2.0 cm in diameter and carries a 650-mA current. Find the magnetic field strength (a) at the loop center and (b) on the loop axis, 20 cm from the center. Solution Equation 30-3 gives: (a) at the center, x = 0, B = μ 0 I = 2 a = (4 ! " 10 # 7 N/A 2 )(650 mA) = (2 " 1 cm) = 40.8 T; (b) on the axis, x = 20 cm, B = 1 2 0 Ia 2 ( x 2 + a 2 ) ! 3 = 2 = 1 2 0 I (1 cm) 2 " [(20 cm) 2 + (1 cm) 2 ] ! 3 = 2 = 5.09 nT. Problem 7. A single-turn current loop carrying 25 A produces a magnetic field of 3.5 nT at a point on its axis 50 cm from the loop center. What is the loop area, assuming the loop diameter is much less than 50 cm? Solution If the radius of the loop is assumed to be much smaller than the distance to the field point ( a ø x = 50 cm), then Equation 30-4 for the field on the axis of a magnetic dipole can be used to find = 2 x 3 B = 0 . The magnetic moment of a single-turn loop is = IA , therefore A = = I = 2 x 3 B = 0 I = (3.5 nT)(50 cm) 3 = (2 " 10 # 7 N/A 2 )(25 A) = 0.875 cm 2 . Problem 8. Two identical current loops are 10 cm in diameter and carry 20-A currents. They are placed 1.0 cm apart, as shown in Fig. 30-47. Find the magnetic field strength at the center of either loop when their currents are in (a) the same and (b) opposite directions. FIGURE 30-47 Problem 8. Solution The magnetic field strength at the center of either loop is the magnitude of the vector sum of the fields due to its own current and the current in the other loop. Equation 30-3 gives B self = 0 I = 2 a and B other = 0 2 = 2( a 2 + x 2 ) 3 = 2 . When the currents are in the same directions, the fields are parallel and B = B self + B other while if the currents are in opposite directions, B = B self ! B other . Numerically, B self = " 10 # 7 )(20)T = (0.1) = 251 T, and B other = B self (1 + x 2 = a 2 ) ! 3 = 2 = (251 T)[1 + (1 cm = 5 cm) 2 ] ! 3 = 2 = 237 so (a) B self + B other = 488 T, and (b) B self ! B other = 14.4 T.
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Problem 10. A single piece of wire is bent so that it includes a circular loop of radius a , as shown in Fig. 30-48. A current I flows in the direction shown. Find an expression for the magnetic field at the center of the loop. FIGURE 30-48 Problem 10. Solution The field at the center is the superposition of fields due to current in the circular loop and straight sections of wire. The former is μ 0 I = 2 a out of the page (Equation 30-3 at x = 0 for CCW circulation), and the latter is 0 I = 2 ! a out of the page (Equation 30-5 at y = a for the very long, straight sections). Their sum is B = (1 + ) 0 I = 2 a out of the page. Problem 12. Four long, parallel wires are located at the corners of a square 15 cm on a side. Each carries a current of 2.5 A, with the top two currents into the page in Fig. 30-49 and the bottom two out of the page. Find the magnetic field at the center of the square. FIGURE 30-49 Problem 12 Solution. Solution The magnetic field from each wire has magnitude 0 I = 2 ( a = 2 ) (from Equation 30-5, with y = a = 2, the distance from a corner of a square of side a to the center). The right-hand rule gives the field direction along one of the diagonals, as shown superposed on Fig. 30-49, such that the fields from currents at opposite corners are parallel. The net field is
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hw7 - Problem 2. A sing le-turn w ire loop is 2.0 cm in d...

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