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

The magnetic field intensity $(H)$ at the centre of a long solenoid carrying a current of $2 \ A$ is found to be $1000 \ A/m$. The number of turns per centimeter of the solenoid is: (Use $\mu_0 = 4 \pi \times 10^{-7} \ T \ m \ A^{-1}$)

$A$ solenoid of length $0.5 \ m$ has a radius of $1 \ cm$ and is made up of $1000$ turns. It carries a current of $10 \ A$. What is the magnitude of the magnetic field inside the solenoid?

$A$ toroid has a core (non-ferromagnetic) of inner radius $25 \; cm$ and outer radius $26 \; cm,$ around which $3500$ turns of a wire are wound. If the current in the wire is $11 \; A$,what is the magnetic field:
$(a)$ outside the toroid,
$(b)$ inside the core of the toroid,and
$(c)$ in the empty space surrounded by the toroid?

If a copper rod carries a direct current,the magnetic field associated with the current will be

Two wires with currents $3 \text{ A}$ and $1.5 \text{ A}$ are enclosed in a circular loop $P$. $A$ third parallel wire with current $1 \text{ A}$ is situated outside the loop as shown. All the wires are perpendicular to the plane of the circular loop. The value of $\oint \vec{B} \cdot d\vec{l}$ around the loop is ($\mu_0$ = permeability of free space) (in $\mu_0$)