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 problemAngular 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 TMy textbook answer is 0.42 T , How is that possible ?!
Muhammed Atef Elkaffas , 5 Years ago
Grade 12
