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A potential difference of 55*10^(-4) Volt is produced across the terminals of a second hand of a clock due to a magnetic field perpendicular to it. Given that length of second hand is 50 cm , calculate magnetic flux density acting on it . This is How I tackled the problem Angular vel.= 2pi / 60 sec. (since it completes one cycle in 60 sec) linear vel. = omega * rad. = 2pi/60 * (0.5) Blv sin90=Epsilon=E.m.f Then , B=Epsilon/(length hand * linear vel)=0.21 T My textbook answer is 0.42 T , How is that possible ?!

A potential difference of 55*10^(-4) Volt is produced across the terminals of a second hand of a clock due to a magnetic field perpendicular to it. Given that length of second hand is 50 cm , calculate magnetic flux density acting on it . 
This is How I tackled the problem
Angular vel.= 2pi / 60 sec. (since it completes one cycle in 60 sec) 
linear vel. = omega * rad. = 2pi/60 * (0.5)
Blv sin90=Epsilon=E.m.f 
Then , B=Epsilon/(length hand * linear vel)=0.21 T
My textbook answer is 0.42 T , How is that possible ?! 

Grade:12

1 Answers

Shivansh
askIITians Faculty 267 Points
2 years ago
Try doing it using the Rate of change of flux concept.
You’ll find that the rate of change of flux will be B.(wR^2/2)
The method you used is wrong because you probably took the velocity of the end of the miniute hand. You cannot do that because its not just the end that is moving, the whole hand is moving.

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