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 current required to be passed through a solenoid of $15\,cm$ length and $60$ turns in order to demagnetize a bar magnet of magnetic intensity $2.4 \times 10^3\,A/m$ is $.........A$.

$A$ solenoid of length $0.5 \ m$ has a radius of $1 \ cm$ and is made up of $1000$ turns. It carries a current of $10 \ A$. What is the magnitude of the magnetic field inside the solenoid?

$A$ magnetic field of $100 \;G$ $(1 \;G = 10^{-4} \;T)$ is required which is uniform in a region of linear dimension about $10 \;cm$ and area of cross-section about $10^{-3} \;m^2$. The maximum current-carrying capacity of a given coil of wire is $15 \;A$ and the number of turns per unit length that can be wound round a core is at most $1000 \;turns \;m^{-1}$. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

$A$ long copper tube of inner radius $R$ carries a current $i$. The magnetic field $B$ inside the tube is

$A$ solenoid has a core made of material with relative permeability $400$. The magnetic field produced in the interior of the solenoid is $1.0 \text{ T}$. The magnetic intensity in $SI$ units is $\alpha \times 10^5$. The value of $\alpha$ is . . . . . . . (Free space permeability $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$.)