Question

A wire carries current I into the junction shown in FIGURE EX29.13. What is the magnetic field at the dot?

Short answer

The magnetic field will be$B_{net}=0$.

Step-by-step solution

Given Information

We have given,

A current carrying wire as shown in the figure.

We have to find the magnetic field at point a.

Simplify

A current-carrying wire produces a magnetic field

${\overrightarrow{B}}_{currentsegment}=\frac{{\mu }_o}{4\pi }\frac{I∆\overrightarrow{s}\times \hat{r}}{r^2}=\frac{{\mu }_o}{4\pi }\frac{I∆s\sin \theta }{r^2}$

From the figure we can see that,

The horizontal wire carrying current with an angle $0^o$with the dot point.

Since$\sin 0^o=0$, the magnetic field at the dot point due to the horizontal current is zero

$B_x=0$

Simplify

When the current will go through the vertical wire, it will divide into two paths up and down with the same magnitude in opposite direction. Both current apply the same magnitude of the magnetic field at the dot point but in opposite direction.

So both magnetic fields will cancel each other.

$B_y=0$

Hence, the total magnetic field at the dot point is zero.

$B_{net}=0$