# Two insulated wires, each 3.00 m long, are taped together to form a two-wire unit that is 3.00 m...

## Question:

Two insulated wires, each 3.00 m long, are taped together to form a two-wire unit that is 3.00 m long. One wire carries a current of 7.00 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 52.0{eq}^o {/eq} relative to a magnetic field whose magnitude is 0.360 T. The magnitude of the net magnetic force experienced by the two-wire unit is 3.13 N. What is the current I?

## Magnetic Force:

When an electric conductor, through which negatively charged particle is flowing is placed in a magnetic region whose value is a function of the magnitude of the current, magnetic field, conductor's length, and angular position of conductor with the magnetic field direction. This force is termed as a magnetic force.

## Answer and Explanation: 1

**Given Data**

- The lengths of the wires is: {eq}{L_w} = 3.0\;{\rm{m}} {/eq}.

- The current in first wire is: {eq}{I_{\max }} = 7.00\;{\rm{A}} {/eq}.

- The angle between wire and magnetic field direction is: {eq}\alpha = 52.0^\circ {/eq}.

- The magnitude of magnetic field is: {eq}{B_m} = 0.360\;{\rm{T}} {/eq}.

- The net force experienced by the wire unit is: {eq}{F_n} = 3.13\;{\rm{N}} {/eq}.

The expression to calculate the force acting on first wire is given by,

{eq}{F_1} = {B_m}{I_{\max }}{L_w}\sin \alpha {/eq}

The expression to calculate the force acting on second wire is given by,

{eq}{F_1} = - {B_m}I{L_w}\sin \alpha {/eq}

Negative sign is used since current is in opposite direction.

The expression to calculate the current in second wire {eq}\left( I \right) {/eq} is given by,

{eq}\begin{align*} {F_n} &= {F_1} + {F_2}\\ {F_n} &= {B_m}{I_{\max }}{L_w}\sin \alpha - {B_m}I{L_w}\sin \alpha \\ {F_n} &= {B_m}{L_w}\left( {{I_{\max }} - I} \right)\sin \alpha \\ I &= {I_{\max }} - \dfrac{{{F_n}}}{{{B_m}{L_w}\sin \alpha }} \end{align*} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} I &= 7.00 - \dfrac{{3.13}}{{0.36 \times 3.0 \times \sin 52^\circ }}\\ & = 3.32\;{\rm{A}} \end{align*} {/eq}

Thus, the current in second wire {eq}\left( I \right) {/eq} is {eq}3.32\;{\rm{A}} {/eq}.

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Chapter 18 / Lesson 7Magnetism and electricity are closely related. When current flows in a wire, a magnetic field is generated. In this lesson, we will explore how magnetic fields can exert forces on parallel current-carrying wires.