The magnitude of the force per unit length acting on a thin wire carrying a current $I=8 \text{ A}$ at a point $O$,if the wire is bent as shown in the figure with a radius $R=10 \pi \text{ cm}$,is (in $\mu \text{N/m}$)

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

The magnetic field at the center of a current-carrying circular loop is $B_{1}$. The magnetic field at a distance of $\sqrt{3}R$ from the center on its axis is $B_{2}$,where $R$ is the radius of the loop. The value of $B_{1} / B_{2}$ is:

$1 \ T = \dots \text{Gauss}$.

$A$ long current-carrying wire produces a magnetic field of $1 \ T$ at a distance of $r$. What will be the magnetic field at distances of $(a)$ $\frac{r}{2}$,$(b)$ $2r$,and $(c)$ $3r$?

An equilateral triangle is made by uniform wires $AB, BC, CA$. A current $I$ enters at $A$ and leaves from the midpoint of $BC$. If the length of each side of the triangle is $L$, the magnetic field $B$ at the centroid $O$ of the triangle is: