$A$ solenoid of length $2 \ m$ carries a current of $20 \ A$. The diameter of the solenoid is $3 \ cm$. If the magnetic field inside the solenoid is $20 \ mT$, then the length of wire forming the solenoid is (assume $\mu_0 = 4 \pi \times 10^{-7} \ H/m$) (in $m$)

Two coaxial solenoids of different radii carry current $I$ in the same direction. Let $\overrightarrow{F_1}$ be the magnetic force on the inner solenoid due to the outer one and $\overrightarrow{F_2}$ be the magnetic force on the outer solenoid due to the inner one. Then

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

The magnetic flux near the axis and inside the air core solenoid of length $80 \text{ cm}$ carrying current $I$ is $1.57 \times 10^{-6} \text{ Wb}$. Its magnetic moment will be (cross-sectional area of a solenoid is very small as compared to its length, $\mu_0 = 4\pi \times 10^{-7} \text{ SI unit}$, $\pi = 3.14$). (in $\text{ Am}^2$)

In a toroid,the magnetic field on the axis is to be calculated given the radius $r = 0.5 \, cm$,current $I = 1.5 \, A$,number of turns $N = 250$,and relative permeability $\mu_r = 700$. Find the magnetic field $B$ (in Tesla). (in $.5$)

An imaginary north pole of $10 \, Am$ is rotating around an infinitely long current-carrying wire at $30 \, \text{revolutions/min}$ on a circular path. If the current in the wire is $5 \, A$, then calculate the work done in one second.