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# A long horizontal wire (P) carries a current$\;50A$ . It is rigidly fixed. Another line wire (Q) is placed directly above and parallel to (P). The weight of wire (Q) is$0.075N/m$ and carries a current of$25A$. Find the position of wire (Q) from (P) so that wire to remains suspended due to the magnetic repulsion(A) $3.33mm$(B) $33.3mm$(C) $334mm$(D) $333mm$

Last updated date: 13th Jun 2024
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Hint We solve this problem by using the formula for magnetic force per unit length of a wire between two current carrying wires. For the wire to remain suspended due to magnetic repulsion the weight and the force due to magnetic repulsion should be equal. We use this condition to find the distance.
Formula use: Force per length between two current carrying wires
$\dfrac{F}{l} = \dfrac{{{\mu _0}2{i_2}{i_1}}}{{4\pi r}}$
Force per length is represented by $\dfrac{F}{l}$
Currents in the two wires is represented by ${i_2}{i_1}$
Distance between the two wires is represented by $r$
Permeability of magnetic field in vacuum ${\mu _0}$

Complete Step by step solution
Since the medium between the wires isn’t mentioned we assume it as vacuum.
Permeability of magnetic field in vacuum is ${\mu _0}$
For the wire to be suspended, weight of the wire acting downwards and the force of repulsion should be equal.
$\dfrac{F}{l} = mg/l$
Weight per unit length of wire is equal to $0.075N/m$
Currents in wires (p) and (Q) are ${i_p} = 50,{i_Q} = 25$respectively
Substituting in the force balance equation
$\dfrac{F}{l} = 0.075$
$0.075 = \dfrac{{4\pi \times {{10}^{ - 7}} \times 2 \times 50 \times 25}}{{4\pi \times r}}$
Force of repulsion is equal to $\dfrac{{4\pi \times {{10}^{ - 7}} \times 2 \times 50 \times 25}}{{4\pi \times r}}$
Here, ${i_p} = 50,{i_Q} = 25$ are the currents in the two wires
$r$ is the distance between the two wires
Solving for the distance
$0.075 = \dfrac{{4\pi \times {{10}^{ - 7}} \times 2 \times 50 \times 25}}{{4\pi \times r}}$
$r = \dfrac{{2 \times 50 \times 25}}{{0.075}} \times {10^{ - 7}}$
$r = 3.33mm$
Hence the distance between the wires should be $r = 3.33mm$

Option (A) $r = 3.33mm$ is the correct answer