The magnetic field due to a straight conductor of uniform cross-section of radius $a$ and carrying a steady current is represented by

The Earth's magnetic field at a given point is $0.5 \times 10^{-5} \, Wb/m^2$. This field is to be annulled by the magnetic induction at the center of a circular conducting loop of radius $5.0 \, cm$. The current required to be flown in the loop is nearly......$A$

$A$ 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):

$A$ long curved conductor carries a current $I$. $A$ small current element of length $dl$ on the wire induces a magnetic field at a point away from the current element. If the position vector between the current element and the point is $\vec{r}$,making an angle $\theta$ with the current element,then the induced magnetic field density $d\vec{B}$ at the point is $(\mu_0 = \text{permeability of free space})$:

$A$ hairpin-like shape as shown in the figure is made by bending a long current-carrying wire. What is the magnitude of the magnetic field at point $P$,which lies at the center of the semicircle?

Two concentric coils each of radius equal to $2\pi \, cm$ are placed at right angles to each other. $3 \, A$ and $4 \, A$ are the currents flowing in each coil respectively. The magnetic induction in $Wb/m^2$ at the centre of the coils will be $(\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m)$.