Find the magnetic field at point $P$ due to a straight line segment $AB$ of length $6\, cm$ carrying a current of $5\, A$. (See figure) $(\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A)$

$A$ long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $B$. It is then bent into a circular loop of $n$ turns. The magnetic field at the centre of the coil for the same current will be

$A$ conducting wire carrying current $I$ is arranged as shown in the figure. Find the magnetic field at point $O$.

$A$ loop $ABCDA$,carrying current $I=12 \ A$,is placed in a plane,consists of two semi-circular segments of radius $R_1=6 \pi \ m$ and $R_2=4 \pi \ m$. The magnitude of the resultant magnetic field at center $O$ is $k \times 10^{-7} \ T$. The value of $k$ is . . . . . . . (Given $\mu_0=4 \pi \times 10^{-7} \ Tm \ A^{-1}$)

$A$ long straight wire of circular cross-section of radius $a$ is carrying a steady current. The current is distributed uniformly across the cross-section of the wire. The ratio of the magnetic fields at points $0.5a$ and $1.5a$ from the centre of the wire 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 $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$.