The correct curve between the magnetic induction $B$ along the axis of a long solenoid due to current flow $i$ in it and distance $x$ from one end is:

$A$ thick current-carrying cable of radius $R$ carries current $I$ uniformly distributed across its cross-section. The variation of magnetic field $B(r)$ due to the cable with the distance $r$ from the axis of the cable is represented by:

$A$ straight wire of diameter $0.4 \,mm$ carrying a current of $2 \,A$ is replaced by another wire of $0.8 \,mm$ diameter carrying the same current. The magnetic field at distance $R$ from both the wires is $B_1$ and $B_2$ respectively. The relation between $B_1$ and $B_2$ is

$A$ winding wire which is used to frame a solenoid can bear a maximum $10\, A$ current. If the length of the solenoid is $80\, cm$ and its cross-sectional radius is $3\, cm$,then the required length of the winding wire is $(B = 0.2\, T)$.

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 $Am^2$)

$A$ very long straight wire of radius $r$ carries current $I$. The intensity of the magnetic field $B$ at a point,lying at a perpendicular distance $a$ from the axis is proportional to: (where $a < r$)