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A train travels at 108 kmph towards east...

A train travels at 108 kmph towards east. Earth's magnetic field is `B=0.4xx10^(-4)` Tesla and acts downwards at `60^(@)` to the horizontal. Calculate the induced e.m.f. between the ends of a horizontal axis PQ of the train. Given PQ = 2m. Also find which end of PQ is at a higher potential?

Text Solution

Verified by Experts

Induced e.m.f. along PQ is due to the vertical component of B
`B_(v)=0.4xx10^(-4)sin60=sqrt(3)xx0.2xx10^(-4)T`
`PQ=l=2m`
`v=108xx(5)/(18)m//s=30m//s`
Induced e.m.f., `varepsilon=B_(v)lv=sqrt(3)xx0.2xx10^(-4)xx2xx30`
`=12sqrt(3)xx10^(-4)V=20.784xx10^(-4)V=2.078mV`
Using Fleming.s Right Hand Rule, we can see that `varepsilon` acts from Q to P. Hence .P. is at a higher potential.
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