In the previous question,if the current is $i$ and the magnetic field at $D$ has magnitude $B$,

The magnetic field due to a long straight wire carrying a current $I$ is proportional to:

The electric current in a circular coil of $2$ turns produces a magnetic induction $B_{1}$ at its centre. The coil is unwound and is rewound into a circular coil of $5$ turns and the same current produces a magnetic induction $B_{2}$ at its centre. The ratio of $\frac{B_{2}}{B_{1}}$ is:

The magnetic field due to a current-carrying square loop of side $a$ at a point located symmetrically at a distance of $a/2$ from its centre (as shown in the figure) is:

$A$ non-conducting thin disc of radius $R$ rotates about its axis with an angular velocity $\omega$. The surface charge density on the disc varies with the distance $r$ from the center as $\sigma(r)=\sigma_0\left[1+\left(\frac{r}{R}\right)^\beta\right]$, where $\sigma_0$ and $\beta$ are constants. If the magnetic induction at the center is $B=\left(\frac{9}{10}\right) \mu_0 \sigma_0 \omega R$, the value of $\beta$ is

$A$ 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 $\left( \mu_0 = 4 \pi \times 10^{-7} \, N/A^2 \right)$.