The magnetic field intensity $H$ at the centre of a long solenoid having $n$ turns per unit length and carrying a current $I$,when no material is kept in it,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:

$A$ current $I$ flows along the length of an infinitely long,straight and thin-walled pipe. Then

$A$ toroid has a number of turns per unit length $n$ and current $i$. What is the magnetic field inside the toroid?

$A$ long solenoid is formed by winding $70$ turns $cm^{-1}$. If $2.0\,A$ current flows,then the magnetic field produced inside the solenoid is $.......\times 10^{-4}\,T$ $\left(\mu_0 = 4\pi \times 10^{-7}\,TmA^{-1}\right)$

$A$ long wire carrying a current of $18 \,A$ is kept along the axis of a long solenoid of radius $1 \,cm$. The magnetic field due to the solenoid is $8.0 \times 10^{-3} \,T$. The magnitude of the resultant magnetic field at a point $0.6 \,mm$ from the solenoid axis is (Assume $\mu_0 = 4 \pi \times 10^{-7} \,Tm/A$):