$A$ linear wire carries a current $I$. $A$ single-turn loop is formed from it,and the magnetic field at the center is $B$. If a three-turn loop is formed from the same wire,what will be the magnetic field at the center?

$A$ long straight wire of radius $a$ carries a steady current $I$. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields $B$ and $B'$,at radial distances $\frac{a}{2}$ and $2a$ respectively,from the axis of the wire is

$A$ current $I$ flows in a circular arc of radius $r$ subtending an angle $\theta$ at the centre $O$ as shown in the figure. Find the magnetic field at the centre $O$ of the circle.

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: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)

Find the magnetic field at the point $P$ in the figure. The curved portion is a quarter-circle connected to two long straight wires.

$A$ current-carrying loop $ABCD$ has two circular arcs $AD$ and $BC$ with radii $1 \text{ cm}$ and $2 \text{ cm}$ respectively, as shown in the figure. The two arcs $AD$ and $BC$ subtend a common angle of $30^{\circ}$ at the centre $O$. If the current flowing in the loop is $\frac{1.2}{\pi} \text{ A}$, then the magnitude of the net magnetic field at $O$ is (Given $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$): (in $\mu \text{T}$)