At a distance of $10\, cm$ from a long straight wire carrying current,the magnetic field is $0.04\, T$. At a distance of $40\, cm$,the magnetic field will be....$T$

Consider two infinitely long wires parallel to the $Z$-axis carrying the same current $I$ in the positive $Z$-direction. One wire passes through point $L$ at coordinates $(-1, 1)$ and the other wire passes through point $M$ at coordinates $(-1, -1)$. The resultant magnetic field at the origin $O$ will be

$A$ solenoid of $1000$ turns per metre has a core with relative permeability $500$. Insulated windings of the solenoid carry an electric current of $5 \, A$. The magnetic flux density produced by the solenoid is (permeability of free space $= 4 \pi \times 10^{-7} \, H/m$)

Current $I$ flows through a long conducting wire bent at right angles as shown in the figure. The magnetic field at a point $P$ on the right bisector of the angle $XOY$ at a distance $r$ from $O$ is

$A$ thin ring of $10 \, cm$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40 \pi \, rad \, s^{-1}$ about its axis,perpendicular to its plane. If the magnetic field at its centre is $3.8 \times 10^{-9} \, T$,then the charge carried by the ring is close to $\left( \mu_0 = 4 \pi \times 10^{-7} \, N/A^2 \right)$.

$A$ vertical wire kept in the $Z-X$ plane carries a current from $Q$ to $P$ (see figure). The magnetic field due to the current will have the direction at the origin $O$ along