A current of 10 A is maintained in a long wire directed along the...

The magnetic field due to a long straight current-carrying wire is given by $B = \dfrac{\mu_0 I}{2 \pi r}$

Given $I = 10 \text{ A}$ and perpendicular distance $r = 1 \text{ m}$

Substituting the values:

$B = \dfrac{4 \pi \times 10^{-7} \times 10}{2 \pi \times 1} = 2 \times 10^{-6} \text{ T} = 2 \mu\text{T}$

The direction of the magnetic field is given by the cross product $d\vec{l} \times \vec{r}$

Since the current is along the positive $z$-axis ($d\vec{l} \parallel \hat{k}$) and the position vector is along the positive $x$-axis ($\vec{r} = \hat{i}$), the direction is along $\hat{k} \times \hat{i} = \hat{j}$

Thus, the magnetic field is $2 \mu\text{T}$ along the positive $y$-axis