A long horizontal wire " A^'' carries a current of 50 Amp. It is rigidly fixed. Another small wi

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A long horizontal wire " A^'' carries a current of 50 Amp....

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

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

Key Concepts

Magnetic Field of a Long Straight Conductor

A long, straight wire carrying a current produces a magnetic field that encircles the wire. The magnitude of this magnetic field at a distance d from the wire is given by B = (??I)/(2?d), where ?? is the magnetic constant and I is the current in the wire. This principle comes from Ampere’s Law, which relates current to the generated magnetic field around it.

Force on a Current-Carrying Conductor in a Magnetic Field

A conductor carrying a current in the presence of a magnetic field experiences a force given by F = I (L × B), where I is the current, L is the length vector of the conductor, and B is the magnetic field. The magnitude of the force per unit length is thus F/L = I B sin?, with ? being the angle between the current direction and the magnetic field. When the conductor is aligned appropriately (i.e., perpendicular to B), this force can counteract other forces, such as gravitational weight.

Electromagnetic Equilibrium in Current-Carrying Wires

For a current-carrying wire to remain in equilibrium under the influence of a magnetic force, the magnetic force must balance any other mechanical forces acting on it, such as its weight. In problems of suspended wires, this involves equating the upward magnetic force per unit length to the downward gravitational force per unit length, leading to conditions that determine the required separation between the wires.

Interactions Between Parallel Currents

Parallel currents interact with each other through their magnetic fields. If currents flow in the same direction, the wires attract each other; if they flow in opposite directions, they repel. Understanding this behavior is crucial when analyzing scenarios where magnetic force is used to counteract gravitational force, as the direction of the current in each wire will determine whether the resulting magnetic interaction is attractive or repulsive.

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