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Question
A wire AB is carrying a steady current of 12 A and is lying on the table. Another wire CD carrying 5 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms−2]
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Solution
Force per unit length between the current carrying wires is given as: `F = (μ_0)/(4pi) (2I_1I_2)/r`
Let m be the mass per unit length of wire CD.
As the force balances the weight of the wire.
`therefore F = (μ_0)/(4pi) (2I_1I_2)/r = mg`
Here,m is mass per unit lenght.
`=> 10^-7 xx (2 xx 12 xx 5 )/(1xx 10^-3) = m xx 10`
`=>m= 10^-7 xx (2 xx 12 xx 5 )/(1xx 10^-3) xx 1/10 = 1.2 xx 10^-3 kg m^-1`
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