A rod of length l, along east-west direction is dropped from a height h. If b be the magnetic field due to earth at that place and angle of dip is , then what is the magnitude of induced emf across two ends of the rod when the rod reaches the earth?
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B is the magneetic field and X is the angle of dip
then horizontal magnetic field is BcosX
length of rod is L
velocity of the rod falling is (2gH)½
induced emf is = Blv = Bcosx.L.(2gH)1/2
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The induced emf is Bsinx.L.(2gH)1/2
Length of the rod = l
Let the magnetic field due to earth at the angle of dip is = x.
Therefore,
The horizontal component of the magnetic field will be = B cos x
B cosx will be met at the field accounting for an induced emf in the rod.
Velocity of rod after falling a distance of H = (2gH)1/2
Induced emf
= Blv = Bsinx × l × (2gH)1/2
Thus, the induced emf is Bsinx.L.(2gH)1/2
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