ÌâÄ¿ÄÚÈÝ

10£®ÈçͼËùʾ£¬UÐνðÊôµ¼¹ìÖÃÓڹ⻬¾øÔµµÄË®Æ½ÃæÉÏ£¬µ¼¹ì¼ä¾àΪl=0.4m£¬×ó¶ËÁ¬½ÓµÄµç×èR=0.4¦¸£®½ðÊô°ôab³¤¶ÈL=0.4m£¬µç×èr=0.1¦¸£®ÔÚµ¼¹ì·¶Î§ÄÚÓд¹Ö±Ë®Æ½ÃæµÄ¡¢´Å¸ÐÇ¿¶ÈB=0.1TµÄÔÈÇ¿´Å³¡£¬µ±ÓÃÍâÁ¦Ê¹°ôabÒÔËÙ¶Èv=5m/sÔÈËÙÓÒÒÆÊ±£¬ÊÔÇó£º
£¨1£©Í¨¹ýab°ôµÄµçÁ÷I=0.4A£» 
£¨2£©ab°ôÁ½¶ËµÄµçÊÆ²îUab=0.16V£»
£¨3£©ab°ô¿Ë·þ°²ÅàÁ¦×ö¹¦µÄ¹¦ÂÊP¿Ë=0.08W£®

·ÖÎö £¨1£©ÓÉE=BLvÇó³öµç¶¯ÊÆ£¬ÓÉÅ·Ä·¶¨ÂÉÇó³öµçÁ÷£®
£¨2£©ÓÉÅ·Ä·¶¨ÂÉÇó³öabÁ½¶ËµÄµçÊÆ²î£®
£¨3£©Óɰ²ÅàÁ¦¹«Ê½Çó³ö°²ÅàÁ¦£¬È»ºóÓÉP=FvÇó³ö¹¦ÂÊ£®

½â´ð ½â£º£¨1£©¸ÐÓ¦µç¶¯ÊÆ£ºE=Blv=0.1¡Á0.4¡Á5=0.2V£¬
µçÁ÷£ºI=$\frac{E}{R+r}$=$\frac{0.2}{0.4+0.1}$=0.4A£»
£¨2£©ab°ôÁ½¶ËµÄµçÊÆ²îUab=IR=0.4¡Á0.4=0.16V£»
£¨3£©°²ÅàÁ¦£ºF=BIl=0.1¡Á0.4¡Á0.4=0.016N£¬
¿Ë·þ°²ÅàÁ¦µÄ¹¦ÂÊ£ºP=Fv=0.016¡Á5=0.08W£»
¹Ê´ð°¸Îª£º£¨1£©0.4£»£¨2£©0.16£»£¨3£©0.08£®

µãÆÀ ±¾Ìâ¹Ø¼üÒªÕÆÎÕ¸ÐÓ¦µç¶¯Êƹ«Ê½E=BLv¡¢Å·Ä·¶¨ÂÉ¡¢¹¦Âʹ«Ê½µÈµÈµç´Å¸ÐÓ¦»ù±¾ÖªÊ¶£¬¼´¿ÉÕýÈ·½âÌ⣮Çóab¼äµçÊÆ²îʱҪעÒâ½ðÊô°ôÊǵçÔ´£¬ab¼äµçÊÆ²îÊÇ·¶Ëµçѹ£¬²»ÊǸÐÓ¦µç¶¯ÊÆ£¬Ò²²»ÊÇÄÚµçѹ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø