ÌâÄ¿ÄÚÈÝ

ÈçͼËùʾ£¬Ë®Æ½ÃæÉÏÓÐÁ½¸ùÏà¾à0.5mµÄ×ã¹»³¤µÄƽÐнðÊôµ¼¹ìMNºÍPQ£¬ËüÃǵĵç×è¿ÉºöÂÔ²»¼Æ£¬ÔÚMºÍPÖ®¼ä½ÓÓÐ×èֵΪR=3.0¦¸µÄ¶¨Öµµç×裮µ¼Ìå°ôab³¤l=0.5m£¬Æäµç×èΪr=1.0¦¸£¬Óëµ¼¹ì½Ó´¥Á¼ºÃ£®Õû¸ö×°Öô¦ÓÚ·½ÏòÊúÖ±ÏòÉϵÄÔÈÇ¿´Å³¡ÖУ¬´Å¸ÐӦǿ¶ÈB=0.4T£®ÏÖʹabÒÔv=10m/sµÄËÙ¶ÈÏòÓÒ×öÔÈËÙÔ˶¯£®
£¨1£©abÁ½µãµçÊÆ²îΪ¶àÉÙ£¿
£¨2£©Ê¹ab°ôÏòÓÒÔÈËÙµÄÀ­Á¦FΪ¶àÉÙ£¿
£¨3£©À­Á¦µÄ¹¦ÂÊΪ¶àÉÙ£¿
¾«Ó¢¼Ò½ÌÍø
£¨1£©µç·ÖÐµç¶¯ÊÆ£ºE=Blv=0.4¡Á0.5¡Á10=2V
abÁ½µãµçÊÆ²îΪ£ºUab=
R
R+r
E=
3
3+1
¡Á2=1.5V

£¨2£©µç·ÖеçÁ÷£ºI=
E
R+r
=
2
4
=0.5A

ab°ôÔÈËÙÔ˶¯Ê±À­Á¦Óë°²ÅàÁ¦Æ½ºâ£¬ÔòÓУº
   F=BIl=0.4¡Á0.5¡Á0.5=0.1N
£¨3£©À­Á¦µÄ¹¦ÂÊ£ºP=Fv=0.1¡Á10=1W
´ð£º
£¨1£©abÁ½µãµçÊÆ²îΪÊÇ1.5V£®
£¨2£©Ê¹ab°ôÏòÓÒÔÈËÙµÄÀ­Á¦FΪ0.1N£®
£¨3£©À­Á¦µÄ¹¦ÂÊΪ1W£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø