$A$ solenoid of length $2 \ m$ carries a current of $20 \ A$. The diameter of the solenoid is $3 \ cm$. If the magnetic field inside the solenoid is $20 \ mT$, then the length of wire forming the solenoid is (assume $\mu_0 = 4 \pi \times 10^{-7} \ H/m$) (in $m$)

$A$ solenoid of $1000$ turns per metre has a core of material with relative permeability $400$. The windings of the solenoid are insulated from the core and a current of $2 \ A$ is passed through the solenoid. Then the value of the magnetic intensity inside the solenoid is

$A$ 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 $\,mA$)

Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid decreases.
Reason : Magnetic field energy density is proportional to the square of the current.

To manufacture a solenoid of length $\ell$ and inductance $L$,the length of the thin wire required is (Diameter of the solenoid is very small compared to its length,$\mu_0$ is the permeability of free space).

Give the formula for the magnetic field due to a toroid.