$A$ long solenoid has $n$ turns per meter and current $I \ A$ is flowing through it. The magnetic field at the ends of the solenoid is

The maximum magnetic field produced by a current of $12 \ A$ passing through a copper wire of diameter $1.2 \ mm$ is (in $mT$)

$A$ particle of mass $2.2 \times 10^{-30} \,kg$ and charge $1.6 \times 10^{-19} \,C$ is moving at a speed of $10 \,km/s$ in a circular path of radius $2.8 \,cm$ inside a solenoid. The solenoid has $25 \,turns/cm$ and its magnetic field is perpendicular to the plane of the particle's path. The current in the solenoid is (Take $\mu_0 = 4\pi \times 10^{-7} \,H/m$) (in $\,mA$)

$A$ very long straight wire of radius $r$ carries current $I$. The intensity of the magnetic field $B$ at a point,lying at a perpendicular distance $a$ from the axis is proportional to: (where $a < r$)

In a current-carrying long solenoid,the magnetic field produced does not depend upon:

$A$ toroid has a core (non-ferromagnetic) of inner radius $24 \ cm$ and outer radius $26 \ cm$ around which $2000$ turns of a wire are wound. If the current in the wire is $12 \ A$, the magnetic field inside the core of the toroid is: