$A$ solenoid of length $0.5\; m$ has a radius of $1\; cm$ and is made up of $500$ turns. It carries a current of $5\; A$. What is the magnitude of the magnetic field inside the solenoid?

$A$ current $i$ is uniformly distributed over the cross section of a long hollow cylindrical wire of inner radius $R_1$ and outer radius $R_2$. The magnetic field $B$ varies with distance $r$ from the axis of the cylinder as:

$A$ long solenoid with $ 40 $ turns per cm carries a current of $ 1 \,A $. The magnetic energy stored per unit volume is $ J m^{-3} $. (in $\pi$)

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).

The number of turns per unit length of a long solenoid is $10 \, \text{turns/cm}$. If its average radius is $5 \, \text{cm}$ and it carries a current of $10 \, \text{A}$, then the ratio of flux densities obtained at the centre and at the end on the axis will be:

$A$ 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: