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प्रश्न
A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
(i) Write the expression for the instantaneous value of the e.m.f. induced in the wire.
(ii) What is the direction of the e.m.f.?
(iii) Which end of the wire is at the higher potential?
उत्तर
Given
Length of wire = v
Velocity with which it is falling = v
Magnetic field = BH = B Cos θ
Angle between v & B, Φ = 90°
(i) As the wire fall downwards due to this motion e− within the wire also moves downwards and feels a force due to Earth’s magnetic field towards west
`vecF =-evecv xx vecB => -evB`
As electrons moves towards west end of wire they apply opposite or repelling force to new incoming charge. As more and more electron’s come, eventually we achieve an equilibrium situation where no more electrons can come towards west side end. At this particular situation.
eE = −evB
E = −vB
V = −EL
⇒V = + vBL e.m.f. setup within in the wire.
(ii) The direction of the e.m.f. will be the polarity of the rod, which is positive at the east end and negative at the west end.
(iii) As the electrons get accumulated at the west end it would mean that the west end is at negative potential. Which implies east end is at high potential (non zero potential).
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