The magnetic field at the centre of a long solenoid having $400$ turns per unit length and carrying a current $i$ is $6.24 \times 10^{-2} \,T$. The magnetic field at the centre of another long solenoid having $200$ turns per unit length and carrying a current $\frac{i}{2}$ is

If the number of turns per unit length of a solenoid is doubled, what happens to the magnetic field in the solenoid?

$A$ 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?

$A$ current $i$ ampere flows along an infinitely long straight thin-walled tube. The magnetic induction at any point inside the tube is:

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the number of turns per $cm$ is doubled and the current is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the new value of the magnetic field will be:

$A$ long solenoid has $70 \text{ turns } cm^{-1}$ and carries current $I$. An electron moves within the solenoid in a circle of radius $2.5 \text{ cm}$ perpendicular to the solenoid axis. If the speed of the electron is $4.4 \times 10^6 \text{ m s}^{-1}$, then the current $I$ in the solenoid is (Take $\mu_0 = 4 \pi \times 10^{-7} \text{ SI unit}$, mass of electron $= 9 \times 10^{-31} \text{ kg}$, charge of electron $= 1.6 \times 10^{-19} \text{ C}$) (in $\text{ mA}$)