In a toroid,the number of turns per unit length is $1000$ and the current through it is $\frac{1}{4 \pi} \ A$. The magnetic field produced inside (in $Wb/m^2$) will be:

Statement-$1$: Ampere's law can be used to find the magnetic field due to a finite length of a straight current-carrying wire.
Statement-$2$: The magnetic field due to a finite length of a straight current-carrying wire is symmetric about the wire.

The magnetic induction along the axis of a toroidal solenoid is independent of

$A$ 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? $[\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A]$

What are the directions of the magnetic field between and outside a pair of two parallel large sheets carrying currents in the same directions,as illustrated in the figure?

$A$ 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: