The magnetic field intensity inside a current-carrying solenoid is $H = 2.4 \times 10^3 \ A/m$. If the length and the number of turns of the solenoid are $15 \ cm$ and $60$ turns respectively,the current flowing in the solenoid is: (in $A$)

Assertion: The magnetic field produced by a current-carrying solenoid is independent of its length and cross-sectional area.
Reason: The magnetic field inside the solenoid is uniform.

$A$ closely wound solenoid of $80 \,cm$ long has $5$ layers of windings of $400$ turns each. The diameter of the solenoid is $1.8 \,cm$. If the current carried is $8 \,A$, then the magnitude of the magnetic field inside the solenoid near its centre is approximately

$A$ $50 \ cm$ long solenoid has a winding of $400$ turns. What current must pass through it to produce a magnetic field of induction $4 \pi \times 10^{-3} \ T$ at the centre (in $A$)?

Write the equation for the magnetic field at an inside point of a very long solenoid.

$A$ solenoid has a core of a material with relative permeability $400$. The windings of the solenoid are insulated from the core and carry a current of $1 \text{ A}$. If the number of turns is $1000 \text{ per metre}$,the magnetic field $(B)$ is . . . . . . $\text{T}$. (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)