The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is $B_1$. The field at the centre of another coil of radius $2r$ carrying the same current $I$ is $B_2$. The ratio $\frac{B_1}{B_2}$ is

The magnetic field at the centre $O$ in the given figure is

$A$ horizontal overhead power line is at a height of $4\, m$ from the ground and carries a current of $100\, A$ from west to east. The magnetic field directly below it on the ground is $(\mu_0 = 4\pi \times 10^{-7}\, T\, m A^{-1})$.

$A$ current of $10\,A$ is passing through a long wire which has a semicircular loop of radius $20\,cm$ as shown in the figure. The magnetic field produced at the centre $P$ of the loop is:

Two circular coils are made from the same wire,but the radius of the $1^{\text{st}}$ coil is twice that of the $2^{\text{nd}}$ coil. If the magnetic field at their centers is the same,then the ratio of the potential difference applied across them is ($1^{\text{st}}$ to $2^{\text{nd}}$ coil).

$A$ long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having $n$ turns. If the same current is passed in both the cases, the ratio of magnetic field produced at the centre for one turn to that of $n$ turns is