A railway track running north-south has two parallel rails $1.0\;m$ apart. Calculate the value of induced emf between the rails when a train passes at a speed of $90km/h$. The horizontal component of earth’s field at the place is $0.3\times 10^{-4} Wb/m^{2}$ and angle of dip $60^{\circ}$

Hint: We know that a change in magnetic flux produces a potential difference or the induced emf i.e. electromotive force . we know that the earth has its own magnetic field, thus the motion of the train here is said to induce an emf which can be calculated as shown below.

Complete step-by-step solution:
We know that the magnetic field lines passing per unit area is called the magnetic flux of the system. From the definition of induced emf, we also know that the change in magnetic flux results in a potential difference.
The magnitude of the emf produced is given as $E=Blv$, where $B$ is the magnetic field due to the motion of charges which are separated by a distance $l$ and moving with a velocity $v$.
Here, $B$ is nothing but the earth’s magnetic field. Here, we are considering the moving train to create a change in flux.
Given that The horizontal component of earth’s field due to angle of dip $60^{\circ}$at the place is $0.3\times 10^{-4} Wb/m^{2}$
We know that $B=B_h tan \theta$ then, where $B_h$ is the horizontal component and $\theta$ is the angle of dip.
$\implies B=0.3\times 10^{-4}\times\ tan 60^{\circ}$
$\therefore B=0.3\times \sqrt 3\times 10^{-4}T$
Also given that $v=90km/h =\dfrac{90\times 1000}{60\times 60}=25m/s$ and $l=1m$. Converting both in terms of meters and substituting in the above formula, we have then the emf $E=0.3\times \sqrt 3\times 10^{-4}T\times 25\times 1$
$\therefore E= 1.3\times 10^{-3}V$


Note: Since magnetism and electricity are interrelated, we know that one can induce the other. The induced emf produces a current which follows lenz's law , thus the direction of the induced current is opposing in nature, i.e. tries to oppose or stop the change in flux.