$A$ circular current-carrying coil has radius $R$. At what distance from the centre of the coil on the axis,the magnetic induction will become $\frac{1}{8}$ th of its value at the centre of the coil?

$A$ wire shaped in a regular hexagon of side $2 \text{ cm}$ carries a current of $4 \text{ A}$. The magnetic field at the centre of the hexagon is

Two long straight parallel conductors carry currents $I_1$ and $I_2$ $(I_1 > I_2)$. When the direction of $I_1$ and $I_2$ is the same,the magnetic field intensity midway between the two conductors is $20 \mu T$. If the direction of $I_2$ is reversed,the field intensity becomes $50 \mu T$. The ratio $I_2 / I_1$ is:

Two similar coils, which are separated by $2 \ m$, have a radius of $1 \ m$ and $80$ turns, and share a common axis. Calculate the strength of the magnetic field in microtesla at a point midway between them on their common axis when the current is $0.2 \ A$.

Two identical current-carrying coils with the same center are placed with their planes perpendicular to each other. If current $I = \sqrt{2} \text{ A}$ and the radius of the coil is $R = 1 \text{ m}$,then the magnetic field at the center is equal to (where $\mu_0$ is the permeability of free space).

One of the two identical conducting wires of length $L$ is bent in the form of a circular loop and the other one into a circular coil of $N$ identical turns. If the same current $i$ is passed in both,the ratio of the magnetic field at the center of the loop $(B_L)$ to that at the center of the coil $(B_C)$,i.e.,$\frac{B_L}{B_C}$ will be: