In a region where the Earth's magnetic field is $3 \times 10^{-4} \, T$ with a dip angle $\theta = \tan^{-1}(4/3)$,a metal rod of length $0.25 \, m$ is placed in the North-South direction. If it is moved towards the East with a velocity of $10 \, cm/s$,calculate the induced $emf$ in $\mu V$.

$A$ square loop of area $25 \, cm^2$ has a resistance of $10 \, \Omega$. The loop is placed in a uniform magnetic field of magnitude $40 \, T$. The plane of the loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in $1 \, s$ will be:

$A$ square-shaped wire loop of mass $m$,resistance $R$,and side $a$ moving with speed $v_{0}$,parallel to the $X$-axis,enters a region of uniform magnetic field $B$,which is perpendicular to the plane of the loop. The speed of the loop changes with distance $x$ $(x < a)$ in the field as:

In the given figure,in which direction should the rod be moved to induce an $emf$ between its two ends?

$A$ bar of mass $m$,length $d$,and resistance $R$ slides without friction in a horizontal plane,moving on parallel rails as shown in the figure. $A$ battery that maintains a constant emf $\varepsilon$ is connected between the rails,and a constant magnetic field $\vec{B}$ is directed perpendicularly to the plane of the page. Assuming the bar starts from rest,find the speed at time $t$.

There is a uniform magnetic field $B$ normal to the $xy$ plane. $A$ conductor $ABC$ has length $AB = l_1$,parallel to the $x$-axis,and length $BC = l_2$,parallel to the $y$-axis. $ABC$ moves in the $xy$ plane with velocity $\vec{v} = v_x \hat{i} + v_y \hat{j}$. The potential difference between $A$ and $C$ is proportional to