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  • 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.

Q6.5 A 1.0m  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.

Answers (1)

best_answer

Given 

length of metallic rod :

l=1m

Angular frequency of rotation :

\omega = 400s^{-1}

Magnetic field (which is uniform)

B= 0.5T

Velocity: here velocity at each point of the rod is different. one end of the rod is having zero velocity and another end is having velocity \omega r. and hence we take the average velocity of the rod so,

Average velocity =\frac{0+\omega l}{2}=\frac{\omega l }{2}

Now,

Induce emf 

e=Blv=Bl\frac{wl}{2}=\frac{Bl^2\omega}{2}

e=\frac{0.5*1^2*400}{2}=100V

Hence emf developed is 100V.

Posted by

Pankaj Sanodiya

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