The ratio of the magnetic field inside a solenoid at an axial point well inside and at an axial end point is

$A$ tightly wound long solenoid carries a current of $1.5 \text{ A}$. An electron is executing uniform circular motion inside the solenoid with a time period of $75 \text{ ns}$. The number of turns per metre in the solenoid is . . . . . . .
[Take mass of electron $m_e = 9 \times 10^{-31} \text{ kg}$,charge of electron $|q_e| = 1.6 \times 10^{-19} \text{ C}$,$\mu_0 = 4\pi \times 10^{-7} \text{ N/A}^2$,$1 \text{ ns} = 10^{-9} \text{ s}$]

$A$ long solenoid carrying current $I_1$ produces a magnetic field $B_1$ along its axis. If the current is reduced to $20 \%$ and the number of turns per $cm$ is increased five times,then the new magnetic field $B_2$ is equal to:

$A$ current of $2\, A$ flows in a long,straight wire of radius $2\, mm$. The intensity of the magnetic field on the axis of the wire is:

$A$ solenoid of $1 \ m$ length and $3.55 \ cm$ inner diameter carries a current of $5 \ A$. If the solenoid consists of five closely packed layers each with $700$ turns along its length,then the magnetic field at its centre is (in $mT$)

An infinitely long cylindrical wire of radius $a$ carries a steady current $I$ uniformly distributed across its cross-section. Determine the magnetic field $B$ at a distance $r$ from the axis for: $(a) r > a$,$(b) r = a$,$(c) r < a$,and $(d)$ at the axis $(r = 0)$.