# 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 - Physics

Numerical

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.

#### Solution

Length of the rod, l = 1 m

Angular frequency, ω = 400 rad/s

Magnetic field strength, B = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of lω.

Average linear velocity of the rod, "v" = ("l"omega + 0)/2 = ("l"omega)/2

Emf developed between the centre and the ring,

e = Blv = "Bl"(("l"omega)/2) = ("Bl"^2omega)/2

= (0.5 xx (1)^2 xx 400)/2

= 100 V

Hence, the emf developed between the centre and the ring is 100 V.

Concept: Inductance - Mutual Inductance
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 6 Electromagnetic Induction
Exercise | Q 6.5 | Page 230
NCERT Class 12 Physics Textbook
Chapter 6 Electromagnetic Induction
Exercise | Q 5 | Page 230

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