Question

100匝的线圈在磁感应强度为0.1T的匀强磁场中旋转,转速为50转每秒,面积为500cm,求瞬时感应电压极大值,

Answer 1

f=1/T---T=1/f

w=2*3.14/T

so w=100*3.14=314rad/s

e=Em*sin(wt)

=nBSw*sin(wt)=100*0.1*0.05*314*sin314t

so

when the 314t=3.14/2

then,Em=nBSw= 157V