Physics9C_A_HW9_Soln

# Physics9C_A_HW9_Soln - 28.2 IDENTIFY A moving charge...

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29-1 28.2. IDENTIFY: A moving charge creates a magnetic field as well as an electric field. SET UP: The magnetic field caused by a moving charge is 0 2 sin 4 qv B r µφ π = , and its electric field is 2 0 1 4 e E r = P since q = e . EXECUTE: Substitute the appropriate numbers into the above equations. 71 9 6 0 21 1 2 sin 4 10 T m/A (1.60 10 C)(2.2 10 m/s)sin90 44( 5 . 3 1 0 m ) qv B r ππ −− ×⋅ × × ° == × = 13 T, out of the page. 92 2 1 9 11 1 2 0 1 (9.00 10 m/C) (1 .60 10 ) 5.1 10 /C, 4( 5 . 3 1 0 ) e E r × = × × P toward the electron. EVALUATE: There are enormous fields within the atom! 28.8. IDENTIFY: Both moving charges create magnetic fields, and the net field is the vector sum of the two. The magnetic force on a moving charge is mag sin Fq v B φ = and the electrical force obeys Coulomb’s law. SET UP: The magnetic field due to a moving charge is 0 2 sin 4 qv B r µ = . EXECUTE: (a) Both fields are into the page, so their magnitudes add, giving B = B e + B p = 0 22 ep sin90 4 ev ev rr  + °   B = () 19 0 11 1.60 10 845,000 m/s 4 (5.00 10 ) (4.00 10 ) ×+ ×× B = 1.39 × 10 –3 T = 1.39 mT, into the page. (b) Using 0 2 sin 4 qv B r = , where r = 41 nm and = 180° arctan(5/4) = 128.7°, we get 9 4 41 0 T m / A ( 1 . 6 1 0 )(845,000 m/s)sin128.7 ? .58 10 T, 4 (4 0 ) B × ° × × into the page. (c) 19 4 17 mag sin90 (1.60 10 C)(845,000 m/s)(2.58 10 T) 3.48 10 N, v B = × × = × in the + x direction. 2 1 9 2 12 elec 0 (9.00 10 ) (1/ 4 ) / 5.62 10 N, 0 ) Fe r × = × × P at 51.3° below the + x- axis measured clockwise. EVALUATE: The electric force is much stronger than the magnetic force. 28.23. IDENTIFY: The net magnetic field at the center of the square is the vector sum of the fields due to each wire. SET UP: For each wire, 0 2 I B r = and the direction of B r is given by the right-hand rule that is illustrated in Figure 28.6 in the textbook. EXECUTE: (a) and (b) B = 0 since the magnetic fields due to currents at opposite corners of the square cancel. (c) The fields due to each wire are sketched in Figure 28.23. 0 cos45 cos45 cos45 cos 45 4 cos45 4 cos45 2 abcd a μI BB B B B B π r = ° = ° . (10 cm) (10 cm) 10 2 cm 0.10 2 m r =+ = = , so 7 4 (4 10 T m A) (100 A) 4 cos 45 4.0 10 T, to the left. 2 (0.10 2 m) π B = × EVALUATE: In part (c), if all four currents are reversed in direction, the net field at the center of the square would be to the right.

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29-2 Chapter 29 Figure 28.23 28.30. IDENTIFY: The magnetic field at the center of a circular loop is 0 2 I B R µ = . 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 1 2 that of a full loop. SET UP: Since the straight sections produce no field at P , the field at P is 0 4 I B R = . EXECUTE: 0 4 I B R = . The direction of B r is given by the right-hand rule: B r is directed into the page. EVALUATE: For a quarter-circle section of wire the magnetic field at its center of curvature is 0 8 I B R = .
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Physics9C_A_HW9_Soln - 28.2 IDENTIFY A moving charge...

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