An equilateral triangle of side '$a$' carries a current '$i$'. What is the magnetic field at point '$P$' (a vertex of the triangle)?

The magnetic field at the center of a circular coil carrying current '$I$' for a single turn of a given length of wire is '$B$'. The same wire is bent into a circular coil having two turns. When the same current '$I$' passes through it, the value of the magnetic field becomes:

Two insulated circular loops $A$ and $B$ of radius '$a$' carry a current of '$I$' in the directions as shown in the figure. The magnitude of the magnetic induction at the centre $O$ will be:

$A$ straight wire of diameter $0.5\, mm$ carrying a current of $1\, A$ is replaced by another wire of $1\, mm$ diameter carrying the same current. The strength of the magnetic field at a point far away is

$A$ rod with a circular cross-section area $2 \text{ cm}^2$ and length $40 \text{ cm}$ is wound uniformly with $400$ turns of an insulated wire. If a current $0.4 \text{ A}$ flows in the wire winding, the total magnetic flux produced inside the winding is $4\pi \times 10^{-6} \text{ Wb}$. The relative permeability of the rod is (Given permeability of vacuum $\mu_0 = 4\pi \times 10^{-7} \text{ N m A}^{-2}$)

An element $\Delta l = \Delta x \hat{i}$ is placed at the origin and carries a large current $I = 10 \; A$ (Figure). What is the magnetic field on the $y$-axis at a distance of $0.5 \; m$? Given $\Delta x = 1 \; cm$.