$A$ current flows through a circular coil of radius $R$. Find the distance from the center on the axis where the magnetic field is $\frac{1}{8}$th of the magnetic field at the center.

The magnetic field due to a current-carrying circular loop of radius $5 \ cm$ at a point on the axis at a distance of $12 \ cm$ from the centre is $250 \ \mu T$. The magnetic field at the centre of the loop is (in $\mu T$)

As shown in the figure,a long straight conductor with a semicircular arc of radius $R = \frac{\pi}{10} \, m$ is carrying a current $I = 3 \, A$. The magnitude of the magnetic field at the center $O$ of the arc is $........... \mu T$. (The permeability of the vacuum $\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A$)

$A$ wire carrying current $i$ is shaped as shown. Section $AB$ is a quarter circle of radius $r$. The magnetic field at the center $C$ is directed:

In the diagram,$I_{1}$ and $I_{2}$ are the strengths of the currents in the loop and straight conductor,respectively. Given $OA = AB = R$. The net magnetic field at the centre $O$ is zero. Then the ratio of the currents in the loop and the straight conductor is:

$A$ tightly wound $100$ turns coil of radius $10 \,cm$ carries a current of $7 \,A$. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 \pi \times 10^{-7} \,SI$ units):