Three long wires all lie in an xyplane parallel to the xaxis. They are spaced equally,10 cm apart. The two outer wires each carry a current of 5.0 Ain the positive xdirection. What is the magnitude of the force on a3.0 m section of either of the outer wires if the current in the center wire is 3.2 A(a) in the positive xdirection and (b) in the negative xdirection?

Short Answer

Expert verified
  1. The force positive x-direction is, 1.7×10-4N.
  2. The force negative x-direction is, 2.1×10-5N.

Step by step solution

01

Identification of the given data

  1. Current through outer wires is,i=5.0A,iT=5.0A=ib=i.
  2. Current through the central wire is, ic=3.2A.
  3. The length of the section of the top wire is, L=3.0m.
  4. The separation between any two wires is,d=10×10-2m.
02

Understanding the concept

Parallel wires carrying current in the same direction attract each other, and the parallel wires carrying current in the opposite direction repel each other.

Formula:

  1. Force on the current-carrying wire placed in the magnetic field, F=iL×B.
  2. Force on the wires of lengthLcarrying currenti1and i2and separated by distance, F=(μ0i1i2L2πd).
03

(a) Calculate the magnitude of the force on the 3.0 m section of the top wire if the current in the central wire is in the positive x direction.

The magnitude of the force on a 3.0 m section of top wire if the current in the central wire is in the positive x direction, is given by,

FTc=μ0iciTL2πd

This is a force on T due to C. Here T and C stand for top wire and central wire.

Similarly, we can write,

FTb=μ0ibiTL2π2d

This is a force on T due to b, and b stands for a bottom wire.

Hence, the total force acting on T is,

F=FTc+FTb=iTL·μ02πdic+ib2F=2μ0iTL4πdic+ib2

Substitute all the values in the above equation.

F=2×4π×10-7T.m/A×5.0A×3.0m4π×10×10-2m3.2A+5.0A2=1.7×10-4N

All wires carry current in the same direction, so they attract each other. The top wire is pulled down by the other two wires.

Hence the force positive x-direction islocalid="1662852182289" 1.7×10-4N.

04

(b) Calculate the magnitude of the force on the 3.0 m section of the top wire if the current in the central wire is in the negative x direction.

The magnitude of the force on a 3.0 m section of top wire if the current in the central wire is in the negative x direction, is given by

FTb=μ0ibiTL2π2d

This is a force on T due to b, and it is attractive.

FTc=μ0iciTL2πd

This is a force on T due to C, and it is repulsive.

Hence, the total force acting on T is

F=FTc-FTbF=2μ0iTL4πdic-ib2

Substitute all the values in the above equation.

F=2×4π×10-7T.m/A×5.0A×3.0m4π×10×10-2m3.2A-5.0A2=2.1×10-5N

The top wire is pushed by the central wire in the upward direction and pulled by the bottom wire in the downward direction.

Hence the force negative x-direction is 2.1×10-5N.

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