As shown in Fig. 31-5, two long parallel wires are 10 cm apart in air and carry currents of 6.0

step 1 In this given problem there are two straight and parallel current carrying conductors. The gap b

As shown in Fig. 31-5, two long parallel wires are 10 cm...

As shown in Fig. $31-5$, two long parallel wires are $10 \mathrm{~cm}$ apart in air and carry currents of $6.0 \mathrm{~A}$ and $4.0 \mathrm{~A}$. Find the force on a $1.0$ $\mathrm{m}$ length of wire $D$ if the currents are $(a)$ parallel and $(b)$ antiparallel. (a) This is the situation shown in $\underline{\text { Fig. }} 31-5 .$ The field at wire $D$ due to wire $C$ is directed into the page and has the value $$B=\frac{\mu_{0} I}{2 \pi r}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(6.0 \mathrm{~A})}{2 \pi(0.10 \mathrm{~m})}=1.2 \times 10^{-5} \mathrm{~T}$$ The force on $1 \mathrm{~m}$ of wire $D$ due to this field is $F_{M}=I L B \sin \theta=(4.0 \mathrm{~A})(1.0 \mathrm{~m})\left(1.2 \times 10^{-5} \mathrm{~T}\right)\left(\sin 90^{\circ}\right)=48 \mu \mathrm{N}$ The right-hand rule applied to wire $D$ tells us the force on $D$ is toward the left. The wires attract each other. (b) If the current in $D$ flows in the reverse direction, the force direction will be reversed. The wires will repel each other. The force per meter of length is still $48 \mu \mathrm{N}$.

As shown in Fig. $31-5$, two long parallel wires are $10 \mathrm{~cm}$ apart in air and carry currents of $6.0 \mathrm{~A}$ and $4.0 \mathrm{~A}$. Find the force on a $1.0$ $\mathrm{m}$ length of wire $D$ if the currents are $(a)$ parallel and $(b)$ antiparallel.
(a) This is the situation shown in $\underline{\text { Fig. }} 31-5 .$ The field at wire $D$ due to wire $C$ is directed into the page and has the value
$$B=\frac{\mu_{0} I}{2 \pi r}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(6.0 \mathrm{~A})}{2 \pi(0.10 \mathrm{~m})}=1.2 \times 10^{-5} \mathrm{~T}$$
The force on $1 \mathrm{~m}$ of wire $D$ due to this field is
$F_{M}=I L B \sin \theta=(4.0 \mathrm{~A})(1.0 \mathrm{~m})\left(1.2 \times 10^{-5} \mathrm{~T}\right)\left(\sin 90^{\circ}\right)=48 \mu \mathrm{N}$
The right-hand rule applied to wire $D$ tells us the force on $D$ is toward the left. The wires attract each other.
(b) If the current in $D$ flows in the reverse direction, the force direction will be reversed. The wires will repel each other. The force per meter of length is still $48 \mu \mathrm{N}$.

Key Concepts

Interaction Between Parallel Currents

This concept covers how two current-carrying wires interact with each other magnetically. When currents in two parallel wires are in the same direction, the wires attract each other due to the overlapping magnetic fields, while currents in opposite directions lead to a repulsive force. This interaction is a direct consequence of the magnetic fields generated by the currents and is important in the design and analysis of electrical circuits and devices.

Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the magnetic field created by a current-carrying conductor or the direction of the force on a conductor in a magnetic field. By orienting your right hand with your thumb in the direction of the current, your curled fingers indicate the direction of the magnetic field. This rule is essential for correctly assigning vector directions in electromagnetic problems.

Magnetic Field of a Long Straight Current-Carrying Wire

This concept describes how a steady current produces a magnetic field around a long, straight conductor. The field at a distance r from the wire is given by B = ??I/(2?r), where ?? is the permeability of free space and I is the current. This expression is derived from either Ampère's law or the Biot–Savart law and is fundamental in understanding how magnetic fields arise from moving charges.

Magnetic Force on a Current-Carrying Conductor

This concept explains how a current-carrying conductor placed in a magnetic field experiences a force. The force is calculated using F = ILB sin?, where I is the current, L is the length of the conductor in the field, B is the magnetic field strength, and ? is the angle between the current direction and the magnetic field. This force, known as the Lorentz force, is key to numerous applications in electromagnetism and electrical engineering.

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