$A$ conducting rod $PQ$ of length $l = 5 \ m$ is oriented as shown in the figure. It is moving with a velocity $\vec{V} = (2 \ m/s) \hat{i}$ without any rotation in a uniform magnetic field $\vec{B} = (3 \hat{j} + 4 \hat{k}) \ T$. The induced $Emf$ in the rod is...........$V$.

$A$ rod of length $l$,mass $m$,and resistance $R$ slides without friction down parallel conducting rails as shown in the figure. The rails are connected together at the bottom. The plane of the rails makes an angle $\theta$ with the horizontal and a uniform vertical magnetic field $B$ exists throughout the region. Then the induced $emf$ in the loop,at the time the rod slides down with a speed $v$,is

$A$ vertical rod of length $l$ is moved with constant velocity $v$ towards the east. If the vertical component of the Earth's magnetic field is $B$ and the angle of dip is $\theta$,then the induced $emf$ in the rod is:

$A$ wheel with ten metallic spokes,each $0.50\,m$ long,is rotated with a speed of $120\,rev/min$ in a plane normal to the Earth's magnetic field at the place. If the magnitude of the field is $0.40\,G$,the induced $emf$ between the axle and the rim of the wheel is equal to:

$A$ 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:

$A$ simple pendulum with a bob of mass $m$ and a conducting wire of length $L$ swings under gravity through an angle $2\theta$. The Earth's magnetic field component in the direction perpendicular to the swing is $B$. The maximum potential difference induced across the pendulum is