The magnetic field intensity inside a current-carrying solenoid is $H = 2.4 \times 10^3 \ A/m$. If the length and the number of turns of the solenoid are $15 \ cm$ and $60$ turns respectively,the current flowing in the solenoid is: (in $A$)

The maximum magnetic field produced by a current of $12 \ A$ passing through a copper wire of diameter $1.2 \ mm$ is (in $mT$)

$A$ straight wire of diameter $0.4 \,mm$ carrying a current of $2 \,A$ is replaced by another wire of $0.8 \,mm$ diameter carrying the same current. The magnetic field at distance $R$ from both the wires is $B_1$ and $B_2$ respectively. The relation between $B_1$ and $B_2$ is

Using Ampere's circuital law,derive the expression for the magnetic field at a perpendicular distance $r$ from an infinitely long current-carrying wire. Explain it.

The magnetic field inside a long solenoid is:

$A$ toroid has a core of inner radius $r_1$ and outer radius $r_2$,around which $N$ turns of wire are wound. If the current in the wire is $I$,then the magnetic field inside the toroid is $(\mu_0 = \text{permeability of free space})$