$A$ thin circular frame of radius $a$ is made of insulating material. $A$ square loop is constructed within it. If the loop carries current $I$,then the magnetic induction at the geometrical centre $O$ will be:

Two parallel conducting wires of equal length are placed distance $d$ apart and carry currents $I_1$ and $I_2$ respectively in opposite directions. The resultant magnetic field at the midpoint of the distance between both the wires is

The Biot-Savart law for a moving point charge $q$ with velocity $\vec{v}$ is given by $\vec{B} = \frac{\mu_0}{4\pi} \frac{q(\vec{v} \times \vec{r})}{r^3}$. This indicates that the magnetic field $\vec{B}$ produced by an electron moving with velocity $\vec{v}$ is such that:

$A$ horizontal overhead power line carries a current of $90 \text{ A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5 \text{ m}$ above the line?

The magnetic field at the centre of a circular coil of radius $r$,due to current $I$ flowing through it,is $B$. The magnetic field at a point along the axis at a distance $r/2$ from the centre is

Two long parallel wires $X$ and $Y$,separated by a distance of $6 \text{ cm}$,carry currents of $5 \text{ A}$ and $4 \text{ A}$,respectively,in opposite directions as shown in the figure. The magnitude of the resultant magnetic field at point $P$,which is at a distance of $4 \text{ cm}$ from wire $Y$,is $x \times 10^{-5} \text{ T}$. The value of $x$ is . . . . . . .
Take the permeability of free space as $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$.