A railway track running North South has two parallel rails 1.0m apart. Calculate the value of induced e.m.f. between the rails when a train passes at a speed of `90 kmh(-1)`. Horizontal component of earth’s field at that place is `0.3xx1O(-4)` `Wbm(-2)` and angle of dip is `60^@`.

Explanation

Distance between the rails =l = 1 m
Induced emf =e = ?
`v = 90 km//h =(90xx1000)/(3600)` m/s
v=25 m/s
Horizontal component of earth.s magnetic field `=B_(h)=0 cdot 3 xx10^(-4)" wbm"^(-2)`
Angle of dip `= delta =60^(@)`
As the train is moving horizontally, it will intercept the vertical component of earth.s field `(B_(1))`
`B_(v) =B_(h) tan delta =0 cdot 3 xx10^(-4) xx tan 60^(@)`
`=0 cdot 3 xx 10^(-4) xx sqrt(3)`
`=0 cdot 3 xx 10^(-4) xx 1 cdot 732`
`B_(v) =0.5196 xx 10^(-4) T`
`e=B_(v) I v=0 cdot 5196 xx10^(-4) xx 1xx25`
`e=12 cdot 99 xx10^(-4)=1 cdot 3 xx10^(-3) V =1 cdot 3 mV`