Question

A long wire carrying a $5.0$A current perpendicular to the xy-plane intersects the x-axis at$x=-2.0cm$. A second, parallel wire carrying a $3.0$A current intersects the x-axis at $x=+2.0$cm. At what point or points on the x-axis is the magnetic field zero if

(a) the two currents are in the same direction and

(b) the two currents are in opposite directions?

Short answer

(a) The two currents are in the same direction is $0.5cm$.

(b) The two currents are in opposite directions is $8cm$.

Step-by-step solution

part (a) step 1 : Given information 

We need to find the two currents are in the same direction.

part (a) step 2: SImplify

A current-carrying wire produces a magnetic field and the bot-savart law enables us to calculate the magnitude and direction of this magnetic feild.at any point where the magnetic field due to the segment $∆\overrightarrow{s}$of current-carrying wire is given by equation $\left(29.7\right)$

localid="1650973425969" $B=\frac{{\mu }_o}{2\pi }\frac{I}{r}\left(1\right)$

Where r is the distance between the wire and the point and I is the current of the wire. The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.

Both wires have current in the same direction , so the magnetic field will be zero at a point between them because if we apply the right-handed rule we find the magnetic field will be zero in a distance between them. If we assume that the point is $\mathcal{x}$, the magnetic field at this point equals the difference between both of them and will be

localid="1650973429895" $$\begin{gathered} B_{\mathcal{x}}=\frac{{\mu }_o}{2\pi }\frac{I_1}{r_1}-\frac{{\mu }_o}{2\pi }\frac{I_2}{r_2} \\ 0=\frac{{\mu }_o}{2\pi }\left(\frac{I_1}{r_1}-\frac{I_2}{r_2}\right)\left(2\right) \\ \frac{I_1}{r_1}=\frac{I_2}{r_2} \end{gathered}$$

The first wire is at so its distance is localid="1649189496373" $r_1=2cm+\mathcal{x}$while the second wire is at $-2cm$, so its distance is localid="1649189701249" $r_2=2cm-\mathcal{x}$.Now, we use these expressions into equation localid="1649191938021" $\left(2\right)$to get $\mathcal{x}$by

$$\begin{gathered} \frac{5A}{2cm+\mathcal{x}}=\frac{3A}{2cm-\mathcal{x}} \\ 6cm+3\mathcal{x}=10cm-5\mathcal{x} \\ \mathcal{x}=0.5cm \end{gathered}$$

part (b) step 1: Given information 

We need to find the two currents are in opposite direction .

part (b) step 2: Simplify

Both wires have current in the opposite direction, so the magnetic field will be zero at a point outside them because if we apply the right-hand rule we find that the magnetic field will be zero at a point outside both of them.

The first wire is at $-2cm$,so its distance is distance is $r_1=\mathcal{x}+2cm$while the second wire is at $+2cm$,so its distance islocalid="1649191148066" $r_2=\mathcal{x}-2cm$. Now, we use these expressions into equation $\left(2\right)$to get $\mathcal{x}$as:

$$\begin{gathered} \frac{5A}{\mathcal{x}+2cm}=\frac{3A}{\mathcal{x}-2cm} \\ 3\mathcal{x}+6cm=5\mathcal{x}-10cm \\ \mathcal{x}=8cm \end{gathered}$$