The direction of magnetic lines of force produced by passing a direct current in a conductor is given by

The magnetic field at the centre of a circular coil of radius $R$,carrying current $2 \ A$ is $B_1$. The magnetic field at the centre of another coil of radius $3R$ carrying current $4 \ A$ is $B_2$. The ratio $B_1: B_2$ is

When equal current is passed through two coils,equal magnetic field is produced at their centres. If the ratio of the number of turns in the coils is $8: 15$,then the ratio of their radii will be

The magnetic field at the center of a circular coil of radius $0.1\, m$ having $1000$ turns and carrying a current of $0.1\, A$ is:

Surface charge density on a ring of radius $a$ and width $d$ is $\sigma$ as shown in the figure. It rotates with frequency $f$ about its own axis. Assume that the charge is only on the outer surface. The magnetic field induction at the centre is (Assume that $d \ll a$):

$A$ coil having $2000$ turns is wound tightly in the form of a spiral with inner and outer radii $1 \,cm$ and $3 \,cm$, respectively. When a current $\frac{1}{\pi} \,mA$ passes through the coil, the magnetic field at the centre is calculated to be $K \ln 3 \times 10^{-6} \,T$. The value of $K$ is