$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the number of turns per cm are tripled and the current is made $\left(\frac{1}{4}\right)^{th}$,then the new value of the magnetic field will be:

The magnetic field at point $P$ in the given figure,due to a current $I$ flowing through a cylindrical conductor of radius $R$,is:

There are $50$ turns of a wire in every $1$ $cm$ length of a long solenoid. If $4$ $A$ current is flowing in the solenoid,the approximate value of the magnetic field along its axis at an internal point and at one end will be respectively:

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per cm is halved,the new value of the magnetic field is

$A$ 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 $\mu_0 = 4 \pi \times 10^{-7} \,Tm/A$):

$A$ long,straight wire of radius $a$ carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{3}$ and $2a$ respectively from the axis of the wire is