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 . . . . . . $\pi \ cm$.

$A$ long solenoid is formed by winding $20$ $turns/cm$. The current necessary to produce a magnetic field of $20$ $mT$ inside the solenoid will be approximately ..... $A$ $(\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A})$.

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 $Am^2$)

$A$ solenoid is $1.0 \ m$ long and it has $4250$ turns. If a current of $5.0 \ A$ is flowing through it, what is the magnetic field at its centre? $[\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A]$

$A$ long solenoid of length $L$ has a mean diameter $D$. It has $n$ layers of winding of $N$ turns each. If it carries a current $I$,the magnetic field at its centre will be

$A$ long,straight wire of radius $a$ carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{3}$ and $2a$ respectively from the axis of the wire is