Question

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?

Answer 1

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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cRAZY pRINCE

Answer 2

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