The intensity of the magnetic induction field at the centre of a single turn circular coil of radius $5 \,cm$ carrying a current of $0.9 \,A$ is:

The direction of magnetic lines of force close to a straight conductor carrying current will be

$A$ non-conducting thin disc of radius $R$ rotates about its axis with an angular velocity $\omega$. The surface charge density on the disc varies with the distance $r$ from the center as $\sigma(r)=\sigma_0\left[1+\left(\frac{r}{R}\right)^\beta\right]$, where $\sigma_0$ and $\beta$ are constants. If the magnetic induction at the center is $B=\left(\frac{9}{10}\right) \mu_0 \sigma_0 \omega R$, the value of $\beta$ is

$A$ straight wire carrying a current of $12\; A$ is bent into a semi-circular arc of radius $2.0\; cm$ as shown in Figure $(a)$. Consider the magnetic field $B$ at the centre of the arc.
$(a)$ What is the magnetic field due to the straight segments?
$(b)$ In what way does the contribution to $B$ from the semicircle differ from that of a circular loop and in what way does it resemble?
$(c)$ Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Figure $(b)$?

$A$ long straight conductor is bent into the shape as shown. If it carries a current $i$ and the radius of the circular part is $R$,then find the magnetic field $B$ at the centre of the circular coil.

Two straight infinitely long current-carrying wires are kept along the $z-$axis at the coordinates $(0, a, 0)$ and $(0, -a, 0)$ respectively, as shown in the figure. The current in each wire is equal and directed along the negative $z-$axis (into the plane of the paper). The variation of the magnetic field on the $x-$axis will be approximately: