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A long straight conductor PQ, carrying a...

A long straight conductor PQ, carrying a current of `60A`, is fixed horizontally. Another long conductor XY is kept parallel to PQ at a distance of `4mm`, in air. Conductor XY is free to move and carries a current I. Calcualte the magnitude and direction of current I for which the magnetic repulsion just balances the weight of conductor XY. (Mass per units length for conductor XY is `10^-2kg//m`).

Text Solution

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Here, `I_1=60A, I_2=I_A`,
`r=4mm=4xx10^-3m`,
Mass per unit length of conductor XY,
`m=10^-2kg//m`.
As magnetic repulsion is balancing the weight of conductor XY
so, `(mu_0)/(4pi)(2I_1I_2)/(r)=mg`
or `(10^-7xx2xx60xxI)/(4xx10^-3)=10^-2xx9*8`
or `I=(4xx10^-5xx9*8)/(2xx10^-7xx60)=32*67A`
The current in XY must flow in a direction opposite to that in PQ, because only then the force will be repulsive.
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