Answer

Verified

87.3k+ 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

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

Other Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

A lens forms a sharp image on a screen On inserting class 12 physics JEE_MAIN

Which of the following facts regarding bond order is class 11 chemistry JEE_Main

Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main

Find the number of nitrates which gives NO2gas on heating class 12 chemistry JEE_Main

If temperature of sun is decreased by 1 then the value class 11 physics JEE_Main