# A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current...

## Question:

A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel wire CD which is fixed in a horizontal plane. Show that AB is slightly depressed, it executes simple harmonic motion. Find the period of oscillations.

## Current:

The amount of the charge which flows along the current-carrying wire in the unit time is known as the current. The unit used to express the current is the ohm. The mathematical expression from the ohm's law is {eq}V = IR {/eq}.

## Answer and Explanation:

**Given data**

- The value of the current is {eq}I = 20\;{\rm{A}} {/eq}

- The value of the height is {eq}d = 0.01\;{\rm{m}} {/eq}

The expression for the time period for an oscillation is,

{eq}T = 2\pi \sqrt {\dfrac{\omega }{g}} ....................(I) {/eq}

The expression for the magnetic force between two wires is,

{eq}{F_m} = \dfrac{{{\mu _o}{i_1}{i_2}}}{{2\pi d}} {/eq}

Differentiate the above equation with respect to x .

{eq}\begin{align*} \dfrac{{d{F_m}}}{{dx}} &= \dfrac{{ - {\mu _o}{i_1}{i_2}}}{{2\pi {x^2}}}dx\\ d{F_m} &= \left( {\dfrac{{ - mg}}{d}} \right)dx \end{align*} {/eq}

From above expression its is concluded that,

{eq}d{F_m} \propto - dx {/eq}

The above expression shows Ab is slightly depressed.

The expression for the acceleration of the wire is,

{eq}a = \dfrac{{d{F_m}}}{m} {/eq}

Substitute the value in the above equation.

{eq}a = - \left( {\dfrac{g}{d}} \right)dx {/eq}

Substitute the value in the equation (I).

{eq}T = 2\pi \sqrt {\dfrac{d}{g}} {/eq}

Substitute the numeric data in the above equation.

{eq}\begin{align*} T &= 2\pi \sqrt {\dfrac{{0.01}}{{9.8}}} \\ &= 0.2\;\sec \end{align*} {/eq}

Thus, the value of the time period of the oscillation is {eq}0.2\;\sec {/eq}

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