# Visualize please refer to figure ex3334 solve we can

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Visualize: Please refer to Figure EX33.34. Solve: We can use the right-hand rule to determine which current direction experiences an upward force. The current being from right to left, the force will be up if the magnetic field B G points out of the page. The forces will balance when F = ILB = mg ( )( ) ( )( ) 3 2 2.0 10 kg 9.8 m/s 0.131 T 1.5 A 0.10 m mg B IL × = = = Thus B G = (0.131 T, out of page).

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33.35. Model: Assume that the magnetic field is uniform over the 10 cm length of the wire. Force on top and bottom pieces will cancel. Visualize: Please refer to Figure EX33.35. The figure shows a 10-cm-segment of a circuit in a region where the magnetic field is directed into the page. Solve: The current through the 10-cm-segment is 15 V 5 A 3 I R = = = Ω E and is flowing down . The force on this wire, given by the right-hand rule, is to the right and perpendicular to the current and the magnetic field. The magnitude of the force is F = ILB = (5 A)(0.10 m)(50 mT) = 0.025 N Thus F G = (0.025 N, right).
33.36. Model: Two parallel wires carrying currents in opposite directions exert repulsive magnetic forces on each other. Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other. Visualize: Please refer to Figure EX33.36. Solve: The magnitudes of the various forces between the parallel wires are ( ) ( )( )( ) 7 4 0 1 2 2 on 1 2 on 3 3 on 2 1 on 2 2 10 T m/A 0.50 m 10 A 10 A 5.0 10 N 2 0.02 m LI I F F F F d μ π × = = = × = = = ( ) ( )( )( ) 7 4 0 1 3 3 on 1 1 on 3 2 10 T m/A 0.50 m 10 A 10 A 2.5 10 N 2 0.04 m LI I F F d μ π × = = = × = Now we can find the net force each wire exerts on the other as follows: ( ) ( ) ( ) 4 4 4 4 on 1 2 on 1 3 on 1 ˆ ˆ ˆ 5.0 10 N 2.5 10 N 2.5 10 N 2.5 10 N, up F F F j j j = + = × + − × = × = × G G G ( ) ( ) 4 4 on 2 1 on 2 3 on 2 ˆ ˆ 5.0 10 N 5.0 10 N 0 N F F F j j = + = − × + + × = G G G ( ) ( ) ( ) 4 4 4 4 on 3 1 on 3 2 on 3 ˆ ˆ ˆ 2.5 10 N 5.0 10 N 2.5 10 N 2.5 10 N, down F F F j j j = + = × + − × = − × = × G G G

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33.37. Model: Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other. Visualize: Please refer to Figure EX33.37. The current in the circuit on the left is I 1 and has a clockwise direction. The current in the circuit on the right is I 2 and has a counterclockwise direction. Solve: Since 1 9 V 2 4.5 A, I = Ω = the force between the two wires is ( ) ( )( ) 7 2 5 0 1 2 2 10 T m/A 0.10 m 4.5 A 5.4 10 N 2 0.0050 m I LI I F d μ π × = × = = I 2 = 3.0 A 9 V 3.0 3.0 A R = = Ω
33.38. Model: The torque on a current loop is due to the magnetic field. Visualize: Solve: (a) We can use the right-hand rule to find the force direction on the currents at the top, bottom, front, and back segments of loop 1 and loop 2 in Figure P26.36. We see that top F G and bottom F G are equal and opposite to each other. front F G and back F G are not seen in the figure above because they are perpendicular to the page. But, they are also equal and opposite to each other.

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