$A$ cell is connected between the points $A$ and $C$ of a circular conductor $ABCD$ of center $O$ with angle $AOC = 60^o$. If $B_1$ and $B_2$ are the magnitudes of the magnetic fields at $O$ due to the currents in $ABC$ and $ADC$ respectively,the ratio $B_1/B_2$ is

The magnetic field due to a current-carrying loop of radius $3 \text{ cm}$ at a point on its axis at a distance of $4 \text{ cm}$ from its centre is $54 \mu\text{T}$. Then,the value of the magnetic field at the centre of the loop is: (in $\mu\text{T}$)

To produce a uniform magnetic field directed parallel to a diameter of a cylindrical region,one can use the saddle coils illustrated in the figure. The loops are wrapped over a somewhat flattened tube. Assume the straight sections of wire are very long. The end view of the tube shows how the windings are applied. The overall current distribution is the superposition of two overlapping circular cylinders of uniformly distributed current,one toward you and one away from you. The current density $J$ is the same for each cylinder. The position of the axis of one cylinder is described by a position vector $\vec{a}$ relative to the other cylinder. The magnetic field inside the hollow tube is:

$A$ circular loop of wire and a long straight wire carry currents $I_c$ and $I_e$,respectively,as shown in the figure. Assuming that these are placed in the same plane,the magnetic field will be zero at the centre of the loop when the separation $H$ is:

Two long parallel wires are at a distance $R$ apart. They carry steady equal currents in the same directions as shown in the figure. The ratio of magnetic fields at $A, B$ and $C$ respectively,is

$A$ uniform wire is bent in the form of a circle of radius $R$. $A$ current $I$ enters at $A$ and leaves at $C$ as shown in the figure. If the length $ABC$ is half of the length $ADC$,the magnetic field at the centre $O$ will be