$A$ wire of length $L = 1 \ m$ carries a constant current $I$. The wire is bent to form a single circular loop of radius $R$. The magnetic field at the centre of this loop is $B$. If the same wire is now bent to form a circular loop with $n = 4$ turns,what will be the magnetic field $B'$ at the centre of this new loop?

In the figure shown,there are two semicircles of radii $r_1$ and $r_2$ in which a current $i$ is flowing. The magnetic induction at the centre $O$ will be

Two identical current-carrying coils with the same center are placed with their planes perpendicular to each other. If current $I = \sqrt{2} \text{ A}$ and the radius of the coil is $R = 1 \text{ m}$,then the magnetic field at the center is equal to (where $\mu_0$ is the permeability of free space).

In a wire of radius $1 \ mm$,a steady current of $2 \ A$ uniformly distributed across the cross-section of the wire is flowing. Then the magnetic field at a point $0.25 \ mm$ from the centre of the wire is (in $\mu T$)

The Earth's magnetic induction at a certain point is $7 \times 10^{-5} \ Wb/m^2$. This is to be annulled by the magnetic induction at the center of a circular conducting loop of radius $5 \ cm$. The required current in the loop is......$A$

Two long parallel wires carrying currents $8 \,A$ and $15 \,A$ in opposite directions are placed at a distance of $7 \,cm$ from each other. $A$ point $P$ is equidistant from both the wires such that the lines joining the point to the wires are perpendicular to each other. The magnitude of the magnetic field at point $P$ is $(\sqrt{2}=1.4)$ $(\mu_0=4 \pi \times 10^{-7} \,T \cdot m/A)$.