# A railway track running N-S has 2 parallel rails 1m apart. If the horizontal component of magnetic field​ at that place is 0.3*10^-4 and angle of dip is 60°, then what'll be the value of induced emf b/w the rails when a train passes at a speed of 90km/ h?

Apr 17, 2018

We note that horizontal component of magnetic field​ will not produce any induced emf as the train also moves in the horizontal plane. It is the vertical component of the earth's magnetic field which will produce induced emf.
Vertical component of the earth's magnetic field can be found as

${B}_{v} = {B}_{H} \tan \delta$
where $\delta$ is local dip angle.

Inserting given vales we get

${B}_{v} = 0.3 \times {10}^{-} 4 \times \tan {60}^{\circ}$
$\implies {B}_{v} = 0.52 \times {10}^{-} 4 \setminus W b \setminus {m}^{-} 2$

Induced emf$= {B}_{v} \times \text{distance between the tracks"xx"speed}$
Inserting various values in SI units we get
Induced emf $e = 0.52 \times {10}^{-} 4 \times 1 \times \left(90 \times \frac{1000}{3600}\right)$
Induced emf $e = 1.3 \times {10}^{-} 3 \setminus V$