$A$ straight long wire is carrying current $I$. The ratio of magnetic field due to this wire at perpendicular distance $2 \text{ cm}$ and $5 \text{ cm}$ respectively from the wire is

$A$ non-conducting disc of radius $R$ has a surface charge density which varies with distance from the centre as $\sigma(r) = \sigma_0 \left[1 + \sqrt{\frac{r}{R}}\right]$, where $\sigma_0$ is a constant. The disc rotates about its axis with angular velocity $\omega$. If $B$ is the magnitude of magnetic induction at the centre, then $\frac{B}{\mu_0 \sigma_0 \omega R}$ will be

$A$ current $i$ is flowing in a straight conductor of length $L$. The magnetic induction at a point distant $\frac{L}{4}$ from its centre will be

The current through $ABC$ and $A'B'C'$ is $I$. What is the magnetic field at $P$? Given $BP = PB' = r$ (Here $C', B', P, B, C$ are collinear).

$A$ current $I=5 \text{ A}$ flows along a thin wire shaped as shown in the figure. The radius of the curved part of the wire is $R=100 \text{ mm}$, and the angle $2\phi=90^{\circ}$. The magnitude of the magnetic field at point $O$ is approximately:
$\left[\text{Use, } \frac{\mu_0}{4\pi}=10^{-7} \text{ T m A}^{-1}\right]$ (in $\mu\text{T}$)

$A$ square frame of side $l$ carries a current $i$. The magnetic field at its centre is $B$. The same current is passed through a circular coil having the same perimeter as the square. The field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is