$A$ horizontal overhead powerline is at a height of $5 \,m$ from the ground and carries a current of $150 \,A$ from East to West. The magnetic field directly below it on the ground 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

$A$ moving charge can produce which field? Explain.

$A$ current $I$ flows in an anticlockwise direction in a circular arc of a wire having $\left(\frac{3}{4}\right)^{\text{th}}$ of the circumference of a circle of radius $R$. The magnetic field $B$ at the centre of the circle is $(\mu_0 = \text{permeability of free space})$

$A$ current $I$ flows around a closed path in the horizontal plane as shown in the figure. The path consists of eight arcs with alternating radii $r$ and $2r$. Each segment of arc subtends an equal angle at the common centre $P$. The magnetic field produced by the current path at point $P$ is

Two concentric coplanar circular loops of radii $r_1$ and $r_2$ carry currents $i_1$ and $i_2$ in opposite directions (one clockwise and the other anticlockwise). The magnetic induction at the center of the loops is half that due to $i_1$ alone at the center. If $r_2 = 2r_1$,find the value of $\frac{i_2}{i_1}$.