The magnetic flux passing through the rectangular loop is proportional to:

$A$ current $i$ is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre $M$ of the loop is $\frac{\mu_{0} i}{R}$ times
$(MA=R, MB=2 R, \angle DMA=90^{\circ})$

When a helium nucleus makes a full rotation of a circle of radius $0.8 \,m$ in $2.5 \,s$, then the value of magnetic field $B$ at the centre of the circle will be

$A$ circular coil of wire consisting of $100$ turns,each of radius $8.0 \, cm$,carries a current of $0.40 \, A$. What is the magnitude of the magnetic field $B$ at the centre of the coil?

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:

Write the formula for the magnetic field due to a circular current-carrying loop having $N$ turns and $R$ radius at a point on the axis of the loop at a distance $x$ from the center.