A thin ring of $10 \mathrm{c}\mathrm{m}$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40\mathrm{\pi }\mathrm{ }\mathrm{r}\mathrm{a}\mathrm{d} {\mathrm{s}}^{-1}$ about its axis, perpendicular to its plane. Is the magnetic field its centre is $3.8\times {10}^{-9} \mathrm{T}$ , then the charge carried by the ring is close to $\left({\mathrm{\mu }}_0=4\mathrm{\pi }\times {10}^{-7} \mathrm{N} {\mathrm{A}}^{-2}\right).$

As shown in the figure, two infinitely long, identical wires are bent by ${90}^o$ and placed in such a way that the segments $LP$ and $QM$ are along the $x$ - axis, while segments $PS$ and $QN$ are parallel to the $y$ - axis. If $OP=OQ=4cm,$ and the magnitude of the magnetic field at $O$ is ${10}^{-4} T,$ and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at $O$ will be $\left({\mu }_0=4\pi \times {10}^{-7}N{A}^{-2}\right):$

A current $I$ is flowing along an infinite, straight wire, in the positive $Z-$ direction and the same current is flowing along a similar parallel wire $5 \mathrm{m}$ apart, in the negative $Z-$ direction. A point $P$ is at a perpendicular distance $3 \mathrm{m}$ from the first wire and $4 \mathrm{m}$ from the second. What will be the magnitude of the magnetic field $B$ of $P$?