题干

如图1所示,矩形金属框置于匀强磁场中,ef为一导体棒,可在ab和cd间滑动并接触良好;设磁感应强度为B,ef长为L,在Δt时间内向左匀速滑过距离Δd,由电磁感应定律E=n可知,下列说法正确的是(   )

图1

A:当ef向左滑动时,左侧面积减少L·Δd,右侧面积增加L·Δd,因此E=2BLΔd/Δt

B:当ef向左滑动时,左侧面积减小L·Δd,右侧面积增大L·Δd,互相抵消,因此E=0

C:在公式中,在切割情况下,ΔΦ=B·ΔS,ΔS应是导线切割扫过的面积,因此E=BLΔd/Δt

D:在切割的情况下,只能用E=BLv计算,不能用计算

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C

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