Question

The two insulated wires in $FIGUREP29.42$cross at a ${30}^{\circ }$angle but do not make electrical contact. Each wire carries a $5.0A$current. Points$1$and $2$are each $4.0$cm from the intersection and equally distant from both wires. What are the magnitude and direction of the magnetic fields at points$1$and$2$?

$FIGUREP29.42$ $FIGUREP29.43$

Short answer

(a) $B_1=5.2\times {10}^{-5}T.$

(b)$B_2=0$

Step-by-step solution

Step.1  Given information 

We should find what are the magnitude and direction of the magnetic fields at points $1$ and$2$?

Step.2  solution 

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

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

where r is the distance between the wire and the point and I is the current of the wire. At point $\left(1\right)$, the distance between it and the intersection is $4cm$. So, its distance from the horizontal wire is

$r_1=\left(4cm\right)\sin {75}^{\circ }=3.86cm$

This the same distance from the second wire as we are given

role="math" localid="1649194404108" $r_1=r_2=3.86cm$

if we apply the right-hand rule, we find that the magnetic field at point $1$of each wire has direction out of the magnetic field at point $1$is the summation of both magnetic fields

$B_1=\frac{{\mu }_o}{2\mathrm{\pi }}\frac{I}{r_2}+\frac{{\mu }_o}{2\mathrm{\pi }}\frac{I}{r_2}=2\left(\frac{{\mu }_o}{2\pi }\frac{I}{r_1}\right)$ $\left(2\right)$

Now, we can use the values for $Iand{r}_1$to get the magnetic field at point $1$by

$$\begin{gathered} B_1=2\left(\frac{{\mu }_o}{2\pi }\frac{I}{r_1}\right) \\ =2\left(\frac{\left(4\pi \times {10}^{-7}T·m/A\right)\left(5A\right)}{2\pi \left(3.86\times {10}^{-2}m\right)}\right) \\ =5.2\times {10}^{-5}T \end{gathered}$$

Step.3  solution

(b) The right-hand rule is used to determine the directions of the fields where you point your thumb in the direction of the current while the remaining fingers curl around the wire and point in the direction of the magnetic field.

At point $2$, the horizontal while applies a magnetic field with direction out of the page while the other wire applies a magnetic feild with direction into thepge. As both wires exert the same magnetic field at point $2$is zero

role="math" localid="1649195582814" $B_2=0$