Question
Download Solution PDFThe force between two parallel current carrying conductors placed at a x distance apart and carrying same current I is:
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KVS TGT WET (Work Experience Teacher) Official Paper
Answer (Detailed Solution Below)
Option 2 : \(\rm \frac{2\times 10^{-7}I^2}{ x} \frac{N}{m}\)
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Detailed Solution
Download Solution PDFCONCEPT:
- 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.
From Ampere’s circuital law, the magnitude of the field due to the first conductor can be given by,
Ba = 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 B1 = μ0 I1 I2 L / 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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