Question

A railway track running $N-S$ has two parallel rails $1m$ apart. Calculate the value of induced emf between the rails, when a train passes at a speed of $90km{h}^{-1}$. Horizontal component of earth's field at that space/place is $0.3\times 10^{-4}wb{m}^{-2}$ and angle of dip is $60^{\circ }$.

• A. $1.3\times 10^{-3}V$
• B. $1.3\times 10^{-4}V$
• C. $1.3\times 10^{-2}V$
• D. $1.3\times 10^{-5}V$

Answer 1

Answer: (A)

$V=90km{h}^{-1}\Rightarrow 90\times \frac{5}{18}m/s$
$\Rightarrow 25m/s$
$H=0.3\times 10^{-4}Wb{m}^{-2},\delta =60^{\circ }$
As vertical component intercepted
$\therefore e=Blv\Rightarrow Blv$ $\Rightarrow (H\tan \delta )lv\left(\tan \delta =\frac{B}{H}\right)$
$e=0.3\times 10^{-4}\times \tan 60^{\circ }\times 1\times 25$
$e=0.3\times 10^{-4}\times \sqrt{3}\times 25$
$e=7.5\times 10^{-4}\times 1.71=1.3\times 10^{-3}V$