$A$ long solenoid with $20$ turns per $cm$ is made. To produce a magnetic field of $20$ $mT$ inside the solenoid, the necessary current will be approximately......$A$.

$A$ current of $2\, A$ flows in a long,straight wire of radius $2\, mm$. The intensity of the magnetic field on the axis of the wire is:

An imaginary north pole of $10 \, Am$ is rotating around an infinitely long current-carrying wire at $30 \, \text{revolutions/min}$ on a circular path. If the current in the wire is $5 \, A$, then calculate the work done in one second.

$A$ current $i$ $A$ flows along an infinitely long straight thin-walled tube. The magnetic induction at any point inside the tube is:

$A$ current of $5 \ A$ flows through a toroid having a core of mean radius $20 \ cm$. If $4000$ turns of the conducting wire are wound on the core,then the magnetic field inside the core of the toroid is [permeability of free space $= 4 \pi \times 10^{-7} \ T \cdot m/A$]

$A$ long solenoid with $10.0 \text{ turns/cm}$ and a radius of $8 \text{ cm}$ carries a current of $7 \text{ mA}$. $A$ current-carrying straight conductor is located along the central axis of the solenoid. If the direction of the resulting magnetic field is $60^{\circ}$ to the axial direction at a point $5 \text{ cm}$ from the axis of the solenoid along the radial direction, then the current in the conductor is. [Take $\sqrt{2}=1.4, \sqrt{3}=1.7$] (in $\text{ A}$)