$A$ long straight wire with a circular cross-section having radius $R$ is carrying a steady current $I$. The current $I$ is uniformly distributed across this cross-section. Then the variation of magnetic field due to current $I$ with distance $r$ $(r < R)$ from its centre will be:

$A$ solenoid has a core of a material with relative permeability $400$. The windings of the solenoid are insulated from the core and carry a current of $1 \text{ A}$. If the number of turns is $1000 \text{ per metre}$,the magnetic field $(B)$ is . . . . . . $\text{T}$. (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)

$A$ 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 $1$ and $2$ of the same length are placed such that one is inside the other. The number of turns per unit length are ${n_1}$ and ${n_2}$. The currents ${i_1}$ and ${i_2}$ are flowing in opposite directions. The magnetic field inside the inner coil is zero. This is possible when:

$A$ particle of mass $1 \times 10^{-27} \ kg$ and charge $1 \times 10^{-16} \ C$ enters a uniform magnetic field within a solenoid at a speed of $1000 \ m/s$. The velocity vector makes an angle of $60^{\circ}$ with the axis of the solenoid. The solenoid has $5000$ turns along its length $L$ and carries a current of $5 \ A$. The number of revolutions the particle makes along the helical path within the solenoid by the time it emerges from the solenoid's opposite end is:

$A$ toroidal solenoid with air core has an average radius $R$,number of turns $N$,and area of cross-section $A$. The self-inductance of the solenoid is (Neglect the field variation across the cross-section of the toroid).