Q. 54.2( 24 Votes )

# A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^{–1} about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

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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A rod AB moves with a uniform velocity v in a uniform magnetic field, as shown in the figure.

HC Verma - Concepts of Physics Part 2