The magnetic field at the origin due to the current $I$ flowing in the wire is -

When a helium nucleus covers a circle of radius $0.8 \,m$ in $2 \,s$, find the value of magnetic field $B$ at the centre of the circle.

Two identical long parallel wires carry currents $I_1$ and $I_2$ such that $I_1 > I_2$. When the currents are in the same direction,the magnetic field at a point midway between the wires is $8 \times 10^{-6} \ T$. If the direction of $I_2$ is reversed,the field becomes $3.2 \times 10^{-5} \ T$. The ratio of $I_2$ to $I_1$ is

The magnetic field $d\overrightarrow{B}$ due to a small current element $d\overrightarrow{l}$ at a distance $\overrightarrow{r}$ from an element carrying current $i$ is given by:

Identical currents flow in two perpendicular wires,as shown in the figure. The wires are very close but do not touch. The magnetic field can be zero:

$A$ long current-carrying wire produces a magnetic field of $1 \ T$ at a distance of $r$. What will be the magnetic field at distances of $(a)$ $\frac{r}{2}$,$(b)$ $2r$,and $(c)$ $3r$?