The earth's magnetic induction at a certain point is $7 \times {10^{ - 5}}\,Wb/{m^2}.$ This is to be annulled by the magnetic induction at the centre of a circular conducting loop of radius $5 \,cm$. The required current in the loop is......$A$

Two identical circular wires of radius $20\,cm$ and carrying current $\sqrt{2}\,A$ are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wire is $.............\times 10^{-8}\,T$. (Take $\pi=3.14$ )

Two long straight wires are placed along $x$-axis and $y$-axis. They carry current $I_1$ and $I_2$ respectively. The equation of locus of zero magnetic induction in the magnetic field produced by them is

A tightly wound $100$ turns coil of radius $10 \mathrm{~cm}$ carries a current of $7 \mathrm{~A}$. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 \pi \times 10^{-7} \mathrm{SI}$ units):

A charge $Q$ is uniformly distributed over the surface of nonconducting disc of radius $R$. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity $\omega$. As a result of this rotation a magnetic field ofinduction $B$ is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure

Two circular coils $X$ and $Y$, having equal number of turns, carry equal currents in the same sence and subtend same solid angle at point $O$. If the smaller coil $X$ is midway between $O$ and $Y$, and If we represent the magnetic induction due to bigger coil $Y$ at $O$ as $B_Y$ and that due to smaller coil $X$ at $O$ as $B_X$, then $\frac{{{B_Y}}}{{{B_X}}}$ is