$A$ 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):

The ratio of the magnetic field at the centre of a circular ring of radius $R$ to the magnetic field at a point on its axis at a distance $2\sqrt{2}R$ from its centre 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 $(b)$?

$A$ linear wire carries a current $I$. $A$ single-turn loop is formed from it,and the magnetic field at the center is $B$. If a three-turn loop is formed from the same wire,what will be the magnetic field at the center?

$A$ long solenoid with $1000 \, \text{turns/m}$ has a core material with relative permeability $500$ and volume $10^{3} \, \text{cm}^{3}$. If the core material is replaced by another material having relative permeability of $750$ with the same volume, maintaining the same current of $0.75 \, \text{A}$ in the solenoid, the fractional change in the magnetic moment of the core would be approximately $\left(\frac{x}{499}\right)$. Find the value of $x$.

The magnetic field $d\overrightarrow{B}$ due to a small current element $d\overrightarrow{l}$ at a distance $\overrightarrow{r}$ from an element carrying current $i$ is given by: