Answer

Verified

51.6k+ views

**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**

**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.

Recently Updated Pages

A small illuminated bulb is at the bottom of a tank class 12 physics JEE_Main

A metal wire is subjected to a constant potential difference class 12 physics JEE_main

Deflection in the galvanometer A Towards right B Left class 12 physics JEE_Main

Calculate the resistivity of the material of a wire class 12 physics JEE_Main

When an external voltage V is applied across a semiconductor class 12 physics JEE_Main

When a Voltmeter connected across the terminals of class 12 physics JEE_Main

Other Pages

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main

In Bohrs model of the hydrogen atom the radius of the class 12 physics JEE_Main

When white light passes through a hollow prism then class 11 physics JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main