$A$ closely wound flat circular coil of $25$ turns of wire has a diameter of $10\, cm$ and carries a current of $4\, A$. Determine the magnetic flux density at the centre of the coil.

$A$ particle carrying a charge equal to $1000$ times the charge on an electron is rotating at $1$ rotation per second in a circular path of radius $r \ m$. If the magnetic field produced at the centre of the path is $x$ times the permeability of vacuum $\mu_0$,the radius $r$ in $m$ is: $[e = 1.6 \times 10^{-19} \ C], [x = 2 \times 10^{-16}]$

The magnetic field due to a current-carrying circular loop of radius $3 \ cm$ at a point on the axis at a distance of $4 \ cm$ from the centre is $54 \ \mu T$. What will be its value at the centre of the loop? (in $\mu T$)

$A$ 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 $(b)$?

Given below are two statements:
Statement $I:$ Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element $(Id\vec{l})$ of a current-carrying conductor only.
Statement $II:$ Biot-Savart's law is analogous to Coulomb's inverse square law of charge $q$,with the former being related to the field produced by a vector source,$Id\vec{l}$,while the latter is produced by a scalar source,$q$. In light of the above statements,choose the most appropriate answer from the options given below:

Due to $10 \ A$ of current flowing in a circular coil of $10 \ cm$ radius,the magnetic field produced at its centre is $3.14 \times 10^{-3} \ Wb/m^2$. The number of turns in the coil will be: