The magnetic induction at the centre $O$ is:

$A$ current $i$ is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre $M$ of the loop is $\frac{\mu_{0} i}{R}$ times
$(MA=R, MB=2 R, \angle DMA=90^{\circ})$

In the loop shown,the magnetic induction at the point $O$ is

$A$ long solenoid has $100 \, \text{turns/m}$ and carries current $i$. An electron moves within the solenoid in a circle of radius $2.30 \, \text{cm}$ perpendicular to the solenoid axis. The speed of the electron is $0.046 \, c$ ($c = 3 \times 10^8 \, \text{m/s}$ is the speed of light). Find the current $i$ in the solenoid (approximate). (in $ \text{A}$)

$A$ 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 across its cross-section. The ratio of the magnetic field at $a/2$ and $2a$ is: