Chapter 20 As was demonstrated in class, two wires in proximity to each other an
ID: 2302174 • Letter: C
Question
Chapter 20
As was demonstrated in class, two wires in proximity to each other and carrying current that flows in the same direction will exert attractive forces on each other. The two wires below each carry 5.2 amps of current and 0.6 m long sections of each wire are in proximity to each other. Each wire exerts a 0.4 N force on the other at the separation distance shown. In this context each wire can be viewed as interacting with the magnetic field produced by the other wire at its position. That is, F1 = IL times B21 and F2 = IL times B1, So the force on wire 1 results from the current in wire 1 interacting with the field produced by the current in wire 2. and vice versa. wire 1 wire 2 In which direction do the field vectors produced by the current in wire 2 point, along the length of wire 1? This is the field that interacts with the current in wire 1. Base your answer on the force law F1 = IL times B2 and show specifically how you are determining the direction of B2 along the length of wire 1 (i.e. at wire 1's position). Determine the magnitude of the magnetic field vectors produced by the current in wire 2 along the length of wire 1. Assume the field is uniform over this length. The wires would still attract if wire 1 were now to the right of wire 2. In which direction would the magnetic field vectors produced by wire 2 now point? Explain how you know again basing your answer on the force law F1 = IL times B2.Explanation / Answer
a) The field vector due to wire 2 on wire will be in vertically upward direction that is coming out of the plane of paper. This direction can be found by using the right hand thumb rule. As we see that if we are at wire 1's position, The current in wire two is in + y direction and the position vector of wire 1 with respect to wire 2 is in west direction, so we point our fingers in the direction of current in wire 2, i.e in +y and we face our palm in the direction of position vector that is towards west. Automatically our thumb points vertically upwards giving us the direction of magnetic field.
b) The magnitude of magnetic field by a straight wire at a point p, at a distance r from the wire is given by,
B= u0 I / 2 pie r
Therefore magnitude of magnetic field by wire 2 will be given by B = u0 I / 2 pie r
Force due to this magnetic field = I L B which is given to be 0.4 N
So, 0.4 = I L B
0.4/ 5.2 * 0.6 = B = 0.128 T
c) If the positions of the wires are interchanged, The force will still be attractive between them. The direction of magnetic field due to wire 2 on wire 1 will become opposite i.e vertically downwards (-z) or into the plane of paper. This is because the force on wire 1 to be in the direction of wire 2, The field on wire one must be in -z direction because then only the cross product of the current vector and field vector will point towards the direction of wire 2.
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