Question

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

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