The magnitude of the magnetic field at the centre of an equilateral triangular loop of side $1\,m$ which is carrying a current of $10\,A$ is:......$\mu T$ [Take $\mu _0 = 4\pi \times 10^{-7}\,NA^{-2}$]

$A$ current $I=5 \text{ A}$ flows along a thin wire shaped as shown in the figure. The radius of the curved part of the wire is $R=100 \text{ mm}$, and the angle $2\phi=90^{\circ}$. The magnitude of the magnetic field at point $O$ is approximately:
$\left[\text{Use, } \frac{\mu_0}{4\pi}=10^{-7} \text{ T m A}^{-1}\right]$ (in $\mu\text{T}$)

Two very thin metallic wires placed along $X$ and $Y$-axis carry equal currents as shown in the figure. $AB$ and $CD$ are lines at $45^\circ$ with the axes,with the origin of the axes at $O$. The magnetic field will be zero on the line:

$20 \, A$ current is flowing in a long straight wire. The intensity of the magnetic field at a distance of $20 \, cm$ from the wire will be:

$A$ current-carrying loop $ABCD$ has two circular arcs $AD$ and $BC$ with radii $1 \text{ cm}$ and $2 \text{ cm}$ respectively, as shown in the figure. The two arcs $AD$ and $BC$ subtend a common angle of $30^{\circ}$ at the centre $O$. If the current flowing in the loop is $\frac{1.2}{\pi} \text{ A}$, then the magnitude of the net magnetic field at $O$ is (Given $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$): (in $\mu \text{T}$)

Explain the experiment which demonstrates the magnetic field produced due to a straight long current-carrying wire.