The magnetic induction due to an ideal solenoid is independent of

The magnetic flux near the axis and inside the air core solenoid of length $80 \text{ cm}$ carrying current $I$ is $1.57 \times 10^{-6} \text{ Wb}$. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, $\mu_0 = 4\pi \times 10^{-7} \text{ SI unit}$, $\pi = 3.14$). (in $\text{ Am}^2$)

In the following diagram,there is a straight wire carrying a current $I$. Consider a circular path with radius $R$ near it. If $\vec{B}_T$ is the tangential component of the magnetic field along the circular path,then find the value of the integral $\oint \vec{B}_T \cdot d\vec{l}$.

What is a solenoid? And what is a long solenoid? Explain.

$A$ long solenoid with $10.0 \text{ turns/cm}$ and a radius of $8 \text{ cm}$ carries a current of $7 \text{ mA}$. $A$ current-carrying straight conductor is located along the central axis of the solenoid. If the direction of the resulting magnetic field is $60^{\circ}$ to the axial direction at a point $5 \text{ cm}$ from the axis of the solenoid along the radial direction, then the current in the conductor is. [Take $\sqrt{2}=1.4, \sqrt{3}=1.7$] (in $\text{ A}$)

$A$ solenoid of $1.5 \ m$ length and $4.0 \ cm$ diameter possesses $10 \ turns/cm$. $A$ current of $5 \ A$ is flowing through it. The magnetic induction at the axis inside the solenoid is: