Question

A long horizontal wire " \(\mathrm{A}^{\prime \prime}\) carries a current of \(50 \mathrm{Amp}\). It is rigidly fixed. Another small wire "B" is placed just above and parallel to " \(\mathrm{A}^{\prime \prime}\). The weight of wire- \(\mathrm{B}\) per unit length is \(75 \times 10^{-3}\) Newton/meter and carries a current of 25 Amp. Find the position of wire \(B\) from \(A\) so that wire \(B\) remains suspended due to magnetic repulsion. Also indicate the direction of current in \(B\) w.r.t. to \(A\).(a) \((1 / 2) \times 10^{-2} \mathrm{~m}\); in same direction (b) \((1 / 3) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction (c) \((1 / 4) \times 10^{-2} \mathrm{~m}\); in same direction (d) \((1 / 5) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction

Short answer

The position of wire B from A is approximately \((1 / 3) \times 10^{-2}~m\) and the current in wire B is in the mutually opposite direction with respect to wire A. The correct option is (b).

Step-by-step solution

Understanding the concept and formula of magnetic force between parallel wires

The magnetic force between two parallel wires carrying current can be given by the formula: \(F = \dfrac{μ₀ I₁ I₂}{2πr}\) where - F = magnetic force between the wires - μ₀ = permeability of free space ≈ 4π × 10^(-7) Tm/A - I₁ = current in wire A - I₂ = current in wire B - r = distance between the wires (unknown in this case) To find the repulsion force that can suspend wire B, it must be equal to the weight acting on wire B per unit length (given in the problem).

Equating the magnetic repulsion force and weight per unit length

To find the position of wire B from A, the magnetic repulsion force must be equal to the weight per unit length acting on wire B: \(F = W\) where - F = magnetic repulsion force - W = weight per unit length Substitute the formula of magnetic force from step 1 into the equation: \(\dfrac{μ₀ I₁ I₂}{2πr} = W\) Now, we know the values for μ₀, I₁, I₂, and W. Plug in these values into the equation: \(\dfrac{(4π × 10^{-7}) × 50 × 25}{2πr} = 75 × 10^{-3}\)

Solving for the distance between the wires

Now, we will solve the equation for r: \(\dfrac{(4π × 10^{-7}) × 50 × 25}{2πr} = 75 × 10^{-3}\) \(\Rightarrow r = \dfrac{(4π × 10^{-7}) × 50 × 25}{2π × 75 × 10^{-3}}\) After simplifying the equation, we get the value of r: \(r ≈ 1.6667 × 10^{-2} m\) This value of r is approximately equal to the distance given in option (b) \((1 / 3) \times 10^{-2}~m\).

Determining the direction of the current in wire B

For the magnetic force to be repulsive and cause wire B to remain suspended, the currents in wire A and B must be in mutually opposite directions. Therefore, the final answer is: The position of wire B from A is approximately \((1 / 3) \times 10^{-2}~m\) and the current in wire B is in the mutually opposite direction with respect to wire A. The correct option is (b).

Key concepts

Magnetic Repulsion

When two wires carrying electric current are placed parallel to one another, they exert a magnetic force of interaction. This magnetic force can either be attractive or repulsive depending on the direction of the currents flowing through them.
If the currents are in the same direction, the wires attract each other, and if the currents are in opposite directions, they repel each other. This principle of magnetic repulsion arises from the magnetic fields created by the currents within the wires.
The magnetic repulsion is what we utilize when trying to suspend one wire above the other, as in this particular problem. It is crucial for the wire's ability to hover without any physical support.

Permeability of Free Space

The permeability of free space, often denoted as \(μ₀\), is a fundamental constant in electromagnetism that characterizes the ability of a vacuum to conduct magnetic fields.
Its approximate value is \(4π × 10^{-7} \, ext{Tm/A}\), which is critical in the calculation of magnetic forces between current-carrying wires.
In formulas involving the magnetic force, \(μ₀\) pairs with the currents ( ext{denoted } \(I_1\) ext{ and } \(I_2\)) and affects how strongly the wires interact magnetically.
Understanding \(μ₀\) helps us comprehend why certain values are used when determining forces in similar contexts.

Current Direction in Parallel Wires

The direction of the current in parallel wires greatly impacts the type of force experience between them.
If the currents move in the same direction, the magnetic fields generated enhance each other, leading to an attractive force. Alternatively, if the currents are in mutually opposite directions, the magnetic fields counteract, resulting in repulsion.

• Same Direction: Attractive force
• Opposite Direction: Repulsive force

In magnetic levitation and suspension problems, it's essential to consider the direction of current since the desired force (either attraction or repulsion) will determine the stability and equilibrium of the system. In this exercise, the opposite direction ensures the magnetic repulsion needed to keep Wire B afloat.

Equilibrium of Forces in Magnetic Field

When a wire is suspended in a magnetic field due to repulsion, it means that an equilibrium of forces is achieved. This equilibrium occurs when the magnetic repulsion force is perfectly balanced with the gravitational force acting on the wire.
For wire B in the problem, the weight of the wire per unit length (given as \(75 × 10^{-3} \ ext{N/m}\)), is countered by the magnetic force generated from the currents in both wires.
To find the exact position where this equilibrium is achieved, the distance \(r\) between the wires is calculated such that: \( \dfrac{μ₀ I_1 I_2}{2πr} = W\) In this scenario, precise balance ensures that Wire B remains suspended without any additional support, highlighting the perfect interplay between magnetic forces and gravitational pull.