A straight wire carrying a current of $14\,A$ is bent into a semicircular are of radius $2.2\,cm$ as shown in the figure. The magnetic field produced by the current at the centre $(O)$ of the arc. is $.........\,\times 10^{-4}\, T$

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 :-

$A$ and $B$ are two concentric circular conductors of centre $O$ and carrying currents ${i_1}$ and ${i_2}$ as shown in the adjacent figure. If ratio of their radii is $1 : 2$ and ratio of the flux densities at $O$ due to $A$ and $B$ is $1 : 3$, then the value of ${i_1}/{i_2}$ is

According to Biot-Savart law magnetic field ...... at point of axis of wire.

.......$A$ should be the current in a circular coil of radius $5\,cm$ to annul ${B_H} = 5 \times {10^{ - 5}}\,T$

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