Find the magnetic field at $O$.

$A$ long straight wire of radius $a$ carries a steady current $i$. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at $a/2$ and $2a$ is:

The shape of the magnetic field lines due to an infinitely long,straight current-carrying conductor is:

The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is $B_1$. The field at the centre of another coil of radius $2r$ carrying the same current $I$ is $B_2$. The ratio $\frac{B_1}{B_2}$ is

$A$ very long wire carrying a current $I = 4 \sqrt{2} \, A$ is bent at a right angle. The magnitude of the magnetic field $(|B|)$ at a point $P$ lying on a line perpendicular to the bent wire at a distance $d = 20 \, cm$ from the point of the bending will be (Let $\mu_0 = 4 \pi \times 10^{-7} \, H/m$): (in $\, \mu T$)

When a certain length of wire is turned into one circular loop,the magnetic induction at the centre of the coil due to a current $I$ flowing through it is $B_1$. If the same wire is turned into three loops to make a circular coil,the magnetic induction at the center of this coil for the same current will be: