If we double the radius of a coil keeping the current through it unchanged,then the magnetic field at any point at a large distance from the centre becomes approximately

The current passing through a conducting loop in the form of an equilateral triangle of side $4\sqrt{3} \text{ cm}$ is $2 \text{ A}$. The magnetic field at its centroid is $\alpha \times 10^{-5} \text{ T}$. The value of $\alpha$ is . . . . . . . (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)

$A$ long straight wire, carrying current $I,$ is bent at its midpoint to form an angle of $45^{\circ}.$ The magnetic field induction at point $P,$ at a distance $R$ from the point of bending, is equal to:

$A$ steady current flows in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is $B$. If the same wire is bent into a circular loop of $n$ turns,the magnetic field at the centre of the coil is

$A$ hairpin-like shape as shown in the figure is made by bending a long current-carrying wire. What is the magnitude of the magnetic field at point $P$,which lies at the center of the semicircle?

Two parallel wires of equal lengths are separated by a distance of $3 \ m$ from each other. The currents flowing through the $1^{\text{st}}$ and $2^{\text{nd}}$ wires are $3 \ A$ and $4.5 \ A$ respectively in opposite directions. Find the resultant magnetic field at the midpoint between the wires $(\mu_0 = \text{permeability of free space})$.