If we double the velocity of a rod moving in a uniform magnetic field, then the induced $emf$ will increase by how many times?

$A$ simple pendulum with a bob of mass $m$ and a conducting wire of length $L$ swings under gravity through an angle $\theta$. The component of the Earth's magnetic field in the direction perpendicular to the swing is $B$. The maximum e.m.f. induced across the pendulum is ($g=$ acceleration due to gravity).

$(a)$ Obtain an expression for the mutual inductance between a long straight wire and a square loop of side $a$ as shown in Figure.
$(b)$ Now assume that the straight wire carries a current of $50\; A$ and the loop is moved to the right with a constant velocity, $v=10\; m / s$. Calculate the induced $emf$ in the loop at the instant when $x=0.2\; m$. Take $a=0.1\; m$ and assume that the loop has a large resistance.

$A$ cycle wheel contains $24$ spokes of $0.5 \, m$ length. It is rotated in a horizontal plane with $120 \, \text{revolution/min}$ in the presence of the Earth's magnetic field. If the total magnetic field of the Earth is $10^{-4} \, T$ (given $10^4 \, G = 1 \, T$), then find the dynamic $emf$ induced across the centre and the rim of the wheel (angle of dip is $30^{\circ}$).

Two straight conducting plates form an angle $\theta$ where their ends are joined. $A$ conducting bar in contact with the plates and forming an isosceles triangle with them starts at the vertex at time $t=0$ and moves with constant velocity $\vec{v}$ to the right as shown in the figure. $A$ magnetic field $\vec{B}$ points out of the page. The magnitude of the emf induced at $t=1 \text{ s}$ will be

$A$ conducting rod $AC$ of length $4l$ is rotated about a point $O$ in a uniform magnetic field $\vec B$ directed into the paper. $AO = l$ and $OC = 3l$. Then: