When a certain length of wire is turned into one circular loop,the magnetic induction at the centre of the coil due to a current $I$ flowing through it is $B_1$. If the same wire is turned into three loops to make a circular coil,the magnetic induction at the center of this coil for the same current will be:

$A$ horizontal overhead power line carries a current of $90 \;A$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5 \;m$ below the line?

In the given figure,find the magnetic field at point $O$.

$A$ current-carrying wire in its neighbourhood produces:

$A$ and $B$ are two concentric circular conductors with center $O$,carrying currents $i_1$ and $i_2$ as shown in the figure. If the ratio of their radii is $1:2$ and the ratio of the magnetic flux densities at $O$ due to $A$ and $B$ is $1:3$,then the value of $i_1/i_2$ is:

Two parallel wires in the plane of the paper are at a distance $X_0$ apart. $A$ point charge is moving with speed $u$ between the wires in the same plane at a distance $X_1$ from one of the wires. When the wires carry current of magnitude $I$ in the same direction, the radius of curvature of the path of the point charge is $R_1$. In contrast, if the currents $I$ in the two wires have directions opposite to each other, the radius of curvature of the path is $R_2$. If $\frac{X_0}{X_1}=3$, the value of $\frac{R_1}{R_2}$ is: