If the radius of a coil is halved and the number of turns is doubled,then the magnetic field at the centre of the coil,for the same current,will

$A$ thin ring of $10 \, cm$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40 \pi \, rad \, s^{-1}$ about its axis,perpendicular to its plane. If the magnetic field at its centre is $3.8 \times 10^{-9} \, T$,then the charge carried by the ring is close to $\left( \mu_0 = 4 \pi \times 10^{-7} \, N/A^2 \right)$.

$A$ wire of length $L = 1 \ m$ carries a constant current $I$. The wire is bent to form a single circular loop of radius $R$. The magnetic field at the centre of this loop is $B$. If the same wire is now bent to form a circular loop with $n = 4$ turns,what will be the magnetic field $B'$ at the centre of this new loop?

$A$ coil having $9$ turns carrying a current produces a magnetic field $B_1$ at the centre. Now,the coil is rewound into $3$ turns carrying the same current. Then,the magnetic field at the centre $B_2$ is:

The magnetic field at the centre of a circular coil of radius $r$,through which a current $I$ flows,is:

The magnetic field induction at the centre of a circular coil of radius $5 \,cm$ carrying a current of $0.9 \,A$ is (in $SI$ units) (where $\varepsilon_0$ is the absolute permittivity of air in $SI$ units,and the velocity of light $c = 3 \times 10^8 \,ms^{-1}$):