The magnetic field inside a $200$ turns solenoid of radius $10 \ cm$ is $2.9 \times 10^{-4} \ T$. If the solenoid carries a current of $0.29 \ A$,then the length of the solenoid is . . . . . . $\pi \ cm$.

$A$ closely wound solenoid $80\; cm$ long has $5$ layers of windings of $400$ turns each. The diameter of the solenoid is $1.8\; cm$. If the current carried is $8.0\; A$,estimate the magnitude of $B$ inside the solenoid near its centre.

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 $Am^2$)

$A$ 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 $\pi$)

Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid decreases.
Reason : Magnetic field energy density is proportional to the square of the current.

"On flowing current in a conducting wire, a magnetic field is produced around it." This is a law of