$A$ cylindrical conductor of radius $R$ is carrying a constant current. The plot of the magnitude of the magnetic field,$B$ with the distance $d$,from the centre of the conductor,is correctly represented by the figure:

$A$ 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:

If a long hollow copper pipe carries a current,then the magnetic field is produced:

$A$ solenoid has a core made of material with relative permeability $400$. The magnetic field produced in the interior of the solenoid is $1.0 \text{ T}$. The magnetic intensity in $SI$ units is $\alpha \times 10^5$. The value of $\alpha$ is . . . . . . . (Free space permeability $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$.)

To manufacture a solenoid of length $\ell$ and inductance $L$,the length of the thin wire required is (Diameter of the solenoid is very small compared to its length,$\mu_0$ is the permeability of free space).

The magnetic flux near the axis and inside the air core solenoid of length $60 \, cm$ carrying current '$I$' is $1.57 \times 10^{-6} \, Wb$. Its magnetic moment will be $[\mu_0 = 4 \pi \times 10^{-7} \, SI \, unit$ and cross-sectional area is very small as compared to the length of the solenoid.] (in $Am^2$)