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 ?!
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 ?!
Muhammed Atef Elkaffas , 6 Years ago
Grade 12
