If the radius of a coil is halved and the number of turns is doubled,then the magnetic field at the centre of the coil,for the same current,will

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 $x = \_\_\_$.
$\left[\text{Use } \mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}\right]$

The magnitude of the magnetic field at the center of an equilateral triangular loop of side $1 \ m$ which is carrying a current of $10 \ A$ is $.... \mu T$.

Current $I$ flows through a long conducting wire bent at right angles as shown in the figure. The magnetic field at a point $P$ on the right bisector of the angle $XOY$ at a distance $r$ from $O$ is

$A$ coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $I$ passes through the coil,the magnetic field at the centre is

An electron revolves around a nucleus with rotational frequency $f$ in a circular orbit. Due to this,the magnetic induction produced at the nucleus position is $B$. The radius of the circular orbit is directly proportional to: