$A$ 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:

$A$ rod with a circular cross-section area of $2\,cm^2$ and a length of $40\,cm$ is wound uniformly with $400$ turns of an insulated wire. If a current of $0.4\,A$ flows in the wire windings,the total magnetic flux produced inside the windings is $4\pi \times 10^{-6}\,Wb$. The relative permeability of the rod is (Given: Permeability of vacuum $\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A$)

$A$ long solenoid with $20$ turns per $cm$ is made. To produce a magnetic field of $20$ $mT$ inside the solenoid, the necessary current will be approximately......$A$.

$A$ solenoid $60 \;cm$ long and of radius $4.0\; cm$ has $3$ layers of windings of $300$ turns each. $A$ $2.0 \;cm$ long wire of mass $2.5\; g$ lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of $6.0\; A$ in the wire. What value of current (in $A$) in the windings of the solenoid can support the weight of the wire? $(g=9.8\; m \,s ^{-2})$

Consider a circular current-carrying loop of radius $R$ in the $x-y$ plane with its centre at the origin. Consider the line integral $\Im(L) = \left| \int_{-L}^{L} \vec{B} \cdot d\vec{l} \right|$ taken along the $z$-axis.
$(a)$ Show that $\Im(L)$ monotonically increases with $L$.
$(b)$ Use an appropriate Amperian loop to show that $\Im(\infty) = \mu_0 I$,where $I$ is the current in the wire.
$(c)$ Verify this result directly.
$(d)$ Suppose we replace the circular coil with a square coil of side $R$ carrying the same current $I$. What can you say about $\Im(L)$ and $\Im(\infty)$?

Two wires carrying currents of $5 \ A$ and $2 \ A$ are enclosed in a circular loop as shown in the figure. Another wire carrying a current of $3 \ A$ is situated outside the loop. The value of $\oint \overrightarrow{B} \cdot d\overrightarrow{l}$ around the loop is ($\mu_0 = \text{permeability of free space}$,$d\overrightarrow{l}$ is the length element of the Amperian loop).