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Question
A long straight wire carrying current of 25 A rests on a table as shown in figure. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?
Solution
The force is applied on PQ by a long straight wire carrying a current of 25 A which rests on a table. And the forces which other are repulsive if two straight wires are placed parallel to each other carrying current in opposite direction. Now if the wire PQ is in equilibrium then that repulsive force onPQ must balance its weight.
The magnetic field produced by a long straight wire carrying current of 25 A rests on a table on small wire.
`B = (mu_0I)/(2pih)`
The magnetic force on small conductor is `F = BIl sin θ = BIl`
Force applied on PQ balance the weight of small current carrying wire.
`F = mg = (mu_0I^2l)/(2pih)`
`h = (mu_0I^2l)/(2pimg) = (4pi xx 10^-7 xx (25)^2 xx 1)/(2pi xx 2.5 xx 10^-3 xx 9.8) = 51 xx 10^-4 m`
h = 0.51 cm
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