$A$ long straight wire of radius $a$ carries a steady current $i$. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at $a/2$ and $2a$ is:

$A$ portion of a conductive wire is bent in the form of a semicircle of radius $r$ as shown in the figure. At the centre $O$ of the semicircle,the magnetic induction will be:

$A$ 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 $(b)$?

$A$ circular loop and an infinitely long straight conductor carry equal currents,as shown in the figure. The net magnetic field at the centre of the loop is $B_1$,when the current in the loop is clockwise and $B_2$ when the current in the loop is anti-clockwise. Then $\frac{B_1}{B_2}$ is

Let the current $I$ be associated with an electron of charge $e$ moving in a circular orbit of radius $r$ with speed $v$ around the positively charged nucleus. The ratio $\frac{r}{v}$ is

The magnetic field at the centre of a circular coil of radius $R$ due to current $i$ flowing through it is $B$. The magnetic field at a point along the axis at a distance $R$ from the centre is