Question

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) $333mm$

Answer 1

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
$${\displaystyle \frac{F}{l}}={\displaystyle \frac{{\mu }_{0}2i_2{i}_1}{4\pi r}}$$
Force per length is represented by $${\displaystyle \frac{F}{l}}$$
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.
$${\displaystyle \frac{F}{l}}=mg/l$$
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
$${\displaystyle \frac{F}{l}}=0.075$$
$$0.075={\displaystyle \frac{4\pi \times {10}^{-7}\times 2\times 50\times 25}{4\pi \times r}}$$
Force of repulsion is equal to $${\displaystyle \frac{4\pi \times {10}^{-7}\times 2\times 50\times 25}{4\pi \times r}}$$
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
$$0.075={\displaystyle \frac{4\pi \times {10}^{-7}\times 2\times 50\times 25}{4\pi \times r}}$$
$$r={\displaystyle \frac{2\times 50\times 25}{0.075}}\times {10}^{-7}$$
$r=3.33mm$
Hence the distance between the wires should be $r=3.33mm$

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

Additional Information
Since the force between the two wires is repulsive in nature. The current flowing in the wires should be in the opposite direction. Using the right-hand rule, we can find the direction of current and the magnetic field in a current carrying wire.
The right-hand rule tells us that two wires carrying current in opposition direction have a repulsive force between each other. If the current is in the same direction then the magnetic force is attractive in nature.

Note The weight of the wire given in the question is weight per unit length hence we can directly substitute it in the formula for force per unit length between two current carrying wires. The formula can be derived using Ampere’s Law or Biot-Savart’s Law. For simplicity, it is better to remember the formula.