$A$ closely wound solenoid of length $1 \,m$ has $5$ layers of $500$ turns each. If the magnitude of the magnetic field inside the solenoid near its centre is $4.4 \,mT$, the current carried is: (in $\,A$)

$A$ 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-carrying solenoid is placed vertically and a particle of mass $m$ with charge $Q$ is released from rest. The particle moves along the axis of the solenoid. If $g$ is the acceleration due to gravity, then the acceleration $(a)$ of the charged particle will satisfy:

$A$ long solenoid has $200$ turns per cm and carries a current of $2.5 \, A$. The magnetic field at its centre is (given $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$):

$A$ 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).

$A$ winding wire which is used to frame a solenoid can bear a maximum $10\, A$ current. If the length of the solenoid is $80\, cm$ and its cross-sectional radius is $3\, cm$,then the required length of the winding wire is $(B = 0.2\, T)$.