If current $I$ is flowing in a closed circuit with total resistance $R$,the rate of production of heat energy in the loop as we pull it with a constant speed $V$ is ($L=$ length of conductor,$B=$ magnetic field).

Two long parallel horizontal rails,a distance $l$ apart,each having a resistance $\lambda$ per unit length,are joined at one end by a resistance $R$. $A$ perfectly conducting rod $MN$ of mass $m$ is free to slide along the rails without friction. There is a uniform magnetic field of induction $B$ normal to the plane of the paper and directed into the paper. $A$ variable force $F$ is applied to the rod $MN$ such that,as the rod moves,a constant current $i$ flows through the circuit. The applied force $F$ as a function of distance $x$ of the rod from $R$ is:

$A$ wire of length $L$ is bent into a semicircle and moved with a velocity $v$ in a magnetic field $B$ as shown. What is the induced $emf$ between its two ends?

An aircraft with a wingspan of $40 \text{ m}$ flies with a speed of $1080 \text{ km h}^{-1}$ in the eastward direction at a constant altitude in the northern hemisphere,where the vertical component of the earth's magnetic field is $1.75 \times 10^{-5} \text{ T}$. The emf developed between the tips of the wings is: (in $\text{ V}$)

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 $\mu V$.

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