As shown in the figure,a long straight conductor with a semicircular arc of radius $R = \frac{\pi}{10} \, m$ is carrying a current $I = 3 \, A$. The magnitude of the magnetic field at the center $O$ of the arc is $........... \mu T$. (The permeability of the vacuum $\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A$)

$A$ wire of length $l$ is bent into a circular coil of one turn of radius $R_1$. Another wire of the same material and same area of cross-section and same length is bent into a circular coil of two turns of radius $R_2$. When the same current flows through the two coils, the ratio of magnetic induction at the centres of the two coils is

$A$ circular current-carrying coil has a radius $R$. The distance from the centre of the coil on the axis where the magnetic induction will be $\frac{1}{8}$th of its value at the centre of the coil is:

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

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

$A$ solenoid of $1000$ turns per metre has a core with relative permeability $500$. Insulated windings of the solenoid carry an electric current of $5 \, A$. The magnetic flux density produced by the solenoid is (permeability of free space $= 4 \pi \times 10^{-7} \, H/m$)