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A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0ms^(-1) at right angles to the horizontal component of the earth's magnetic field, 0.30 xx 10^(4) Wbm^(-2). (a) What is the instantaneous value of the emf induced in the wire ? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential ? |
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Answer» Solution :Here l = 10 m, `v=5.0 m s^(-1)` and horizontal component of earth.s MAGNETIC field `B_(H) = 0.30 xx 10^(-4)Wb m^(2)` As `B_(H)`, I and v all the THREE are mutually perpendicular to each other HENCE instantaneous value of the emf induced in the wire has a magnitude `|varepsilon|B_(H).l.v=0.30 xx 10^(-4)xx10xx1.5xx10^(-3)V` (b) In accordance with Fleming.s right hand rule, the direction of induced emf is from west to east. (c) For the purpose of outer circuit the eastern end of the wire is at the HIGHER potential. |
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