Find the magnetic field at the centre $P$ of a square loop of side length $20 \, cm$ carrying a current of $3 \, A$.

$A$ loop carrying current $I$ has the shape of a regular polygon of $n$ sides. If $R$ is the distance from the centre to any vertex,then the magnitude of the magnetic induction vector $B$ at the centre of the loop is

$A$ particle having a charge $50 \ e$ is revolving in a circular path of radius $0.4 \ m$ with $1 \ r.p.s.$ The magnetic field produced at the centre of the circle is $(\mu_0 = 4 \pi \times 10^{-7} \ SI \ units$ and $e = 1.6 \times 10^{-19} \ C)$.

$A$ current $i$ is flowing in a straight conductor of length $L$. The magnetic induction at a point distant $\frac{L}{4}$ from its centre will be

$A$ charged particle $A$ with charge $q = 2 \, C$ has a velocity $v = 100 \, m/s$. When it passes through point $A$ with the velocity in the direction shown in the figure,the strength of the magnetic field at point $B$ due to this moving charge is ....... $\mu T$ $(r = 2 \, m)$.

Two circular coils $P$ and $Q$ of $100$ turns each have the same radius of $\pi \text{ cm}$. The currents in $P$ and $Q$ are $1 \text{ A}$ and $2 \text{ A}$ respectively. $P$ and $Q$ are placed with their planes mutually perpendicular and their centers coinciding. The resultant magnetic field induction at the center of the coils is $\sqrt{x} \text{ mT}$,where $x = \_\_\_$.
$\left[\text{Use } \mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}\right]$