$A$ straight wire of finite length carrying current $I$ subtends an angle of $60^{\circ}$ at point $P$ as shown. The magnetic field at $P$ is

Current $I$ is flowing along the path $ABCDA$ consisting of four edges of a cube (figure $-a$),producing a magnetic field $B_0$ at the centre of the cube. Find the magnetic field $B$ produced at the center of the cube by a current $I$ flowing along the path of the six edges $ABCGHEA$ (figure $-b$).

The magnetic field at a distance $r$ from a long straight wire carrying current $i$ is $0.4 \ T$. The magnetic field at a distance $2r$ is: (in $T$)

$A$ long conducting wire carrying a current $I$ is bent at $120^{\circ}$ (see figure). The magnetic field $B$ at a point $P$ on the angle bisector of the bend at a distance $d$ from the bend is ($\mu_{0}$ is the permeability of free space):

$A$ charge $q$ $C$ moves in a circle at $n$ revolutions per second and the radius of the circle is $r$ $m$. The magnetic field at the centre of the circle is:

The magnitude of the magnetic field at $O$ due to a current-carrying loop as shown in the figure is, where $O$ is the center of two circular portions with radii $1 \, cm$ and $2 \, cm$ respectively. (Take the value of current $I = \frac{1.2}{\pi} \, A$)