In the figure,the magnetic induction at the centre $O$ due to the current in the straight portion $AB$ will be:

Consider a long straight wire of a circular cross-section (radius $a$) carrying a steady current $I$. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire,outside the wire] is half of the maximum possible magnetic field,anywhere due to the wire,will be

Describe the arrangement of iron filings sprinkled around a straight current-carrying wire.

The magnetic field at the centre $O$ due to the given structure is:

$A$ particle is moving with velocity $\overrightarrow{v} = \hat{i} + 3\hat{j}$ and it produces an electric field at a point given by $\overrightarrow{E} = 2\hat{k}$. It will produce a magnetic field at that point equal to (all quantities are in $SI$ units):

$A$ circular arc of radius $r$ carrying current $I$ subtends an angle $\frac{\pi}{8}$ at its centre. The radius of the metal wire is uniform. The magnetic induction at the centre of the circular arc is ($\mu_0 =$ permeability of free space).