Answer :

Given:

Length of the metallic rod = 1.0 m

Angular frequency = 400 rads ^{- 1}

Strength of magnetic field B= 0.5 T

Since it is clear that one end of the metallic rod has zero linear velocity whereas the other end of the rod has linear velocity of lω

∴ The average linear velocity of the metallic rod can be calculated as follows:

v = (lω + 0)/2

⇒ v = lω /2

The e.m.f. developed between the centre and the ring can be calculated as follows:

e = Blv ...(1)

where B is the magnetic field

l is the length of the loop and,

v is the velocity of the rectangular loop

⇒ Substituting the value of v in above equation, we get

e = B × l × (lω/2)

e = (Bl^{2}ω)/2

e = (0.5 T × 1m^{2} × 400 rads ^{- 1})/2

On calculating, we get

e = 100 V

Hence, the emf developed across the ring is 100 V.

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