$A$ 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:

Identify the correct statement.

$A$ 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 $\mu_0 = 4\pi \times 10^{-7} \text{ N m A}^{-2}$)

Two circular coils $1$ and $2$ are made from the same wire. The radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them $V_1 / V_2$,so that the magnetic field at their centre is the same?

$A$ uniform wire is bent in the form of a circle of radius $R$. $A$ current $I$ enters at $A$ and leaves at $C$ as shown in the figure. If the length $ABC$ is half of the length $ADC$,the magnetic field at the centre $O$ will be

$A$ particle having a charge $100 e$ is revolving in a circular path of radius $0.8 \ m$ with $1 \ r.p.s$. The magnetic field produced at the center of the circle in $SI$ unit is ($\mu_0$ is the permeability of vacuum,$e = 1.6 \times 10^{-19} \ C$).