An infinitely long conductor $PQR$ is bent to form a right angle as shown. $A$ current $I$ flows through $PQR$. The magnetic field due to this current at the point $M$ is $H_1$. Now,another infinitely long straight conductor $QS$ is connected at $Q$ so that the current is $I/2$ in $QR$ as well as in $QS$,while the current in $PQ$ remains unchanged. The magnetic field at $M$ is now $H_2$. The ratio $H_1/H_2$ is given by

$A$ circular arc of radius $r$ carrying current $I$ subtends an angle $\frac{\pi}{16}$ at its centre. The radius of the metal wire is uniform. The magnetic induction at the centre of the circular arc is [where $\mu_0$ is the permeability of free space].

In a wire of radius $1 \ mm$,a steady current of $2 \ A$ uniformly distributed across the cross-section of the wire is flowing. Then the magnetic field at a point $0.25 \ mm$ from the centre of the wire is (in $\mu T$)

When equal current is passed through two coils,equal magnetic field is produced at their centres. If the ratio of the number of turns in the coils is $8: 15$,then the ratio of their radii will be

An infinite wire has a circular bend of radius $a$, and carries a current $I$ as shown in the figure. The magnitude of the magnetic field at the origin $O$ is given by:

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