$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is doubled and the number of turns per cm is halved,the new value of the magnetic field is

From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current,the variation of magnetic field in the inside and outside region of the wire is :

$A$ toroid of $n$ turns,mean radius $R$ and cross-sectional radius $a$ carries current $I$. It is placed on a horizontal table taken as $xy$-plane. Its magnetic moment $m$ is:

$A$ 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:

$A$ current of $5 \ A$ flows through a toroid having a core of mean radius $20 \ cm$. If $4000$ turns of the conducting wire are wound on the core,then the magnetic field inside the core of the toroid is [permeability of free space $= 4 \pi \times 10^{-7} \ T \cdot m/A$]

$A$ long,straight wire of radius $a$ carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{3}$ and $2a$ respectively from the axis of the wire is