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# Hwksolns1 - 33.5 The magnetic field is into the page on the left of the wire and it is out of the page on the right of the wire Grab the wire with

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33.5. The magnetic field is into the page on the left of the wire and it is out of the page on the right of the wire. Grab the wire with your right hand in such a way that your fingers point out of the page to the right of the wire. Since the thumb now points down, the current in the wire is down.

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33.6. (a) The force on a charge moving in a magnetic field is on q Fq v B =×= GG G ( qvB sin α , direction of right-hand rule) A positive charge moving to the right with B G into the page gives a force that is up . (b) The direction of the force on a negative charge is opposite the direction determined by the right-hand rule. A negative charge moving up with B G out of the page gives a force to the left.
33.7. (a) The force on a charge moving in a magnetic field is on q Fq v B =×= GG G ( qvB sin α , direction of right-hand rule) A positive charge moving to the right with B G down gives a force into the page . (b) The direction of the force on a negative charge is opposite the direction determined by the right-hand rule. Any charge moving parallel to the field has no force and no deflection.

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33.8. (a) The force on a charge moving in a magnetic field is on q Fq v B =×= GG G ( qvB sin α , direction of right-hand rule) The magnetic field must be in a plane perpendicular to both the v G and F G vectors. Using the right-hand rule for a positive charge moving to the right, the B G field must be out of the page. (b) The direction of the force on a negative charge is opposite the direction determined by the right-hand rule. The force F G on the negative charge is into the page. Since the velocity is to the right, the magnetic field B G must be up .
33.13. The magnet is repelled. The current loop produces a magnetic dipole with the north pole on the left and the south pole on the right. The approaching south pole of the magnet is repelled by the south pole of the current loop.

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33.14. Model: Assume the wires are infinitely long. Visualize: Please refer to Figure EX33.14. Solve: The magnetic field strength at point a is () 00 at a top bottom top bottom 7 0 at a 22 5 at a , out of page , into page 11 1 1 21 0 T m /A1 0 A 2 2.0 cm 4.0 2.0 cm 2.0 10 m 6.0 10 m 6.7 10 T, out of page II BBB dd I B B μμ ππ μ π −− ⎛⎞ =+ = + ⎜⎟ ⎝⎠ ⇒= = × × ⇒=× GGG G At points b and c, 4 at 2 , into page , into page 2.0 10 T, into page B =+= × G 5 at 3 , into page , out of page B = × G
33.16. Solve: (a) The magnetic dipole moment of the superconducting ring is () ( ) 2 23 4 2 1.0 10 m 100 A 3.1 10 A m RI μπ π −− == × = × (b) From Example 33.5, the on-axis magnetic field of the superconducting ring is ( ) ( ) 2 73 2 7 0 ring 3/2 3 2 22 2 10 T m/A 100 A 1.0 10 m 5.0 10 T 2 0.050 m 0.0010 m IR B zR μ × = × ⎡⎤ + + ⎣⎦

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33.19. Visualize: Please refer to Figure EX33.19. Solve: Because B G is in the same direction as the integration path s G from i to f, the dot product of B G and ds G is simply Bds . Hence the line integral () fff 22 iii 0.50 m 0.50 m 0.10 T 2 0.50 m 0.071 T m Bds B d s Bd s B ⋅= = = + = = ∫∫∫ G G
33.20.

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## This note was uploaded on 12/26/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.

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Hwksolns1 - 33.5 The magnetic field is into the page on the left of the wire and it is out of the page on the right of the wire Grab the wire with

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