The current required to be passed through a solenoid of $15\,cm$ length and $60$ turns in order to demagnetize a bar magnet of magnetic intensity $2.4 \times 10^3\,A/m$ is $.........A$.

$A$ thick current-carrying cable of radius $R$ carries current $I$ uniformly distributed across its cross-section. The variation of magnetic field $B(r)$ due to the cable with the distance $r$ from the axis of the cable is represented by:

There are $50$ turns per $cm$ length in a very long solenoid. It carries a current of $2.5 \ A$. The magnetic field at its centre on the axis is . . . . . . $T$.

$A$ solenoid of $1.5 \ m$ length and $4.0 \ cm$ diameter possesses $10 \ turns/cm$. $A$ current of $5 \ A$ is flowing through it. The magnetic induction at the axis inside the solenoid is:

$A$ solenoid of length $0.5 \,m$ has a radius of $1 \,cm$ and is made up of $m$ number of turns. It carries a current of $5 \,A$. If the magnitude of the magnetic field inside the solenoid is $6.28 \times 10^{-3} \,T$, then the value of $m$ 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 $= 4 \pi \times 10^{-7} \ T \cdot m/A$]