The magnetic field at the centre of a circular current carrying-conductor of radius $r$ is $B_c$. The magnetic field on its axis at a distance $r$ from the centre is $B_a$. The value of $B_c : B_a$ will be :-

A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $B$. It is then bent into a circular loop of $n$ $turns$. The magnetic field at the centre of the coil will be

Current $i$ is passed as shown in diagram. If radius of the circle is a, then the magnetic flux density at the centre $P$ will be:

Current $I$ is flowing along the path $ABCDA$ consisting of four edges of a cube (figure $-a$), produces 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 $b$)

A beam of neutrons performs circular motion of radius, $r=1 \,m$. Under the influence of an inhomogeneous magnetic field with inhomogeneity extending over $\Delta r=0.01 \,m$. The speed of the neutrons is $54 \,m / s$. The mass and magnetic moment of the neutrons respectively are $1.67 \times 10^{-27} \,kg$ and $9.67 \times 10^{-27} \,J / T$. The average variation of the magnetic field over $\Delta r$ is approximately ....... $T$

The magnetic field at the origin due to a current element $i\,\overrightarrow {dl} $ placed at position $\vec r$ is

$(i)\,\,\left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{d\vec l\, \times \,\vec r}}{{{r^3}}}} \right)$

$(ii)\,\, - \left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{d\vec l\, \times \,\vec r}}{{{r^3}}}} \right)$

$(iii)\,\left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{\,\vec r \times d\vec l}}{{{r^3}}}} \right)$

$(iv)\, - \left( {\frac{{{\mu _0}i}}{{4\pi }}} \right)\left( {\frac{{\,\vec r \times d\vec l}}{{{r^3}}}} \right)$