如图所示,一矩形线圈在匀强磁场中绕OO′轴匀速转动,磁场方向与转轴垂直.线圈匝数 n=50,电阻r=0.1Ω,

1个回答

  • (1)电动势最大值

    Em=NBSω

    =50×0.2×0.1×0.4×100

    =40V

    (2)求电荷量要用平均电动势

    q=I△t

    =

    NBS

    △t(R+r) △t

    =N

    △BS

    R+r =

    50×0.2×0.1×0.4

    9.9+0.1

    =4×10 -2C

    (3)求消耗的功率时用有效值

    P=I 2R

    = (

    Em

    2 (R+r) ) 2 R

    =79.2W

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