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?

A train travels at 108 kmph towards east. Earth's magnetic field is B=

1.	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?
Answer» Solution :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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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?

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