$A$ coil of $12$ turns is made from a wire of constant length carrying a current $I$. If the number of turns is changed to $3$,what is the percentage change in the magnetic field produced at its centre?

When an electric current is passed through a solenoid,how can we determine which end behaves as a North pole and which as a South pole? Explain.

$A$ non-conducting disc of radius $R$ has a surface charge density which varies with distance from the centre as $\sigma(r) = \sigma_0 \left[1 + \sqrt{\frac{r}{R}}\right]$, where $\sigma_0$ is a constant. The disc rotates about its axis with angular velocity $\omega$. If $B$ is the magnitude of magnetic induction at the centre, then $\frac{B}{\mu_0 \sigma_0 \omega R}$ will be

The magnetic induction at point $O$ for the current-carrying wire shown in the figure is:

$A$ wire in the form of a square of side $a$ carries a current $i$. Then, the magnetic induction at the centre of the square is (Magnetic permeability of free space $= \mu_0$)

$A$ 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?