What current must be passed through a solenoid with $20 \text{ turns/cm}$ to produce a magnetic field of $20 \text{ mT}$ (in $\text{ A}$)? (Given: $\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A}$)

The magnetic flux near the axis and inside the air core solenoid of length $60 \, cm$ carrying current '$I$' is $1.57 \times 10^{-6} \, Wb$. Its magnetic moment will be $[\mu_0 = 4 \pi \times 10^{-7} \, SI \, unit$ and cross-sectional area is very small as compared to the length of the solenoid.] (in $Am^2$)

$A$ solenoid of length $0.5 \ m$ has a radius of $1 \ cm$ and is made up of $250$ turns. It carries a current of $5 \ A$. What is the magnitude of the magnetic field inside the solenoid?

$A$ 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:

$A$ 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 $= 4 \pi \times 10^{-7} \ T \cdot m/A$]