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

Step by step solution

01

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).

02

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: \(\dfrac{(4π × 10^{-7}) × 50 × 25}{2πr} = 75 × 10^{-3}\)

03

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) \((1 / 3) \times 10^{-2}~m\).

04

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).

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