The force between two parallel current carrying conductors placed at a x distance apart and carrying same current I is:

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  1. \(\rm \frac{2\times 10^{-7}I^2}{\pi x} \frac{N}{m}\)
  2. \(\rm \frac{2\times 10^{-7}I^2}{ x} \frac{N}{m}\)
  3. \(\rm \frac{4 \pi \times 10^{-7}I^2}{x} \frac{N}{m}\)
  4. \(\rm \frac{2\pi \times 10^{-7}I^2}{x} \frac{N}{m}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{2\times 10^{-7}I^2}{ x} \frac{N}{m}\)
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Detailed Solution

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CONCEPT:

  • Ampere's Circuital Law: It states the relationship between the current and the magnetic field created by it.
  • This law states that the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.

F1 J.K Madhu 13.05.20 D9

From  Ampere’s circuital law, the magnitude of the field due to the first conductor can be given by,

B= μ0 I1 / 2 π d

The force on a segment of length L of conductor 2 due to conductor 1 can be given as,

F21 = I2 L B= μ0 I1 IL / 2πd

EXPLANATION:

  • From the equation
    \({{\rm{F}}_{21}} = \frac{{{{\rm{\mu }}_{0{\rm{\;}}}}{{\rm{I}}_1}{\rm{\;}}{{\rm{I}}_2}{\rm{\;L}}}}{{2{\rm{\;\pi \;d}}}}\)
  • Since \(I_1=I_2=I\)
  •  
  • F=\(\rm \frac{2\times 10^{-7}I^2}{ x} \frac{N}{m}\)
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