Question

3£®Èçͼ¼×Ëùʾ£¬´øÕýµçµÄÁ£×ÓÒÔˮƽËÙ¶Èv₀´ÓƽÐнðÊô°åMN¼äÖÐÏßOO'Á¬ÐøÉäÈëµç³¡ÖУ¬MN°å¼ä½ÓÓÐÈçͼÒÒËùʾµÄËæʱ¼ät±ä»¯µÄµçѹU_MN£¨MµãµçÊÆ×ÜÊǸßÓÚNµãµÄµçÊÆ£©£¬Á½°å¼äµç³¡¿É¿´×÷ÊǾùÔȵģ¬ÇÒÁ½°åÍâÎ޵糡£®½ôÁÚ½ðÊô°åÓÒ²àÓд¹Ö±Ö½ÃæÏòÀïµÄÔÈÇ¿´Å³¡B£¬CDΪ·Ö½çÏߣ¬EFΪÆÁÄ»£¬½ðÊô°å¼ä¾àΪd£¬³¤Îªl£¬´Å³¡µÄ¿í¶ÈΪd£®ÒÑÖª£ºB=5¡Á10^-3T£¬l=d=0.2m£¬Ã¿¸ö´øÕýµçÁ£×ÓµÄËÙ¶Èv₀=1¡Á10⁵m/s£¬±ÈºÉΪ$\frac{q}{m}={10^8}$C/kg£¬ÖØÁ¦ºöÂÔ²»¼Æ£¬ÊÔÇó£º
£¨1£©´øµçÁ£×ÓÉä³öµç³¡Ê±µÄ×îСËٶȺÍ×î´óËٶȣ»
£¨2£©´øµçÁ£×Ó´òÔÚÆÁÄ»Éϵķ¶Î§£®

Answer 1

·ÖÎö £¨1£©ÓÉÊÜÁ¦·ÖÎöµÃµ½¼ÓËٶȣ¬½ø¶øÇóµÃ×îСËٶȣ»ÔÙ¸ù¾Ý±ß½çÌõ¼þ£¬ÇóµÃ¿ÉÄÜÈ¡µÃµÄ×î´óËٶȼ°ÆäÌõ¼þ£¬½ø¶øÇóµÃ×î´óËٶȣ»
£¨2£©¸ù¾Ý£¨1£©¿ÉÖª½øÈë´Å³¡µÄÁ£×ÓµÄËÙ¶ÈÇé¿ö£¬ÔÙÓÉËÙ¶ÈÇóµÃÁ£×ÓÔڴų¡ÖеÄÔ˶¯Çé¿ö¼°ÆäÓë´òÔÚÆÁÉϵÄλÖùØϵ£¬¾Í¿ÉÇóµÃ´øµçÁ£×Ó´òÔÚÆÁÄ»Éϵķ¶Î§£®

½â´ð ½â£º£¨1£©´øµçÁ£×ÓÔڵ糡ÖÐËùÊܵ糡Á¦ÎªÊúÖ±·½Ïò£¬ËùÒÔ£¬Á£×ÓÔÚˮƽ·½ÏòµÄ·ÖÔ˶¯ÎªÔÈËÙÖ±ÏßÔ˶¯£¬ËùÒÔ£¬Á£×Ó¾­¹ýµç³¡µÄʱ¼ä$t=\frac{l}{{v}_{0}}=2¡Á1{0}^{-6}s$£»
´øµçÁ£×Óͨ¹ýµç³¡ÇøÓòµÄʱ¼ä¼«¶Ì£¬ÔÚÕâ¶Îʱ¼äÄÚ°å¼äµÄµç³¡Ç¿¶È¿ÉÈÏΪ²»±ä£®
µ±µçѹ²îU_MN=0ʱ£¬ÉäÈëµç³¡µÄ´øµçÁ£×Ó²»±»¼ÓËÙ£¬ÑØÖ±ÏßOO'Éä³öµç³¡Ê±µÄËÙ¶È×îС£¬×îСËÙ¶ÈΪ${v_{min}}={v_0}=1¡Á{10^5}m/s$£»
ÉèÁ½°å¼äµçѹU_MN=U_mʱ£¬Á£×ÓÇ¡ºÃ´Ó½ðÊô°åµÄϼ«°å±ßÔµÉä³ö£¬ÔòÁ£×ÓÔ˶¯µÄ¼ÓËÙ¶È$a=\frac{F}{m}=\frac{qE}{m}=\frac{q{U}_{m}}{md}$£¬ËùÒÔ£¬$\frac{d}{2}=\frac{1}{2}¡Á\frac{{q{U_m}}}{md}{£¨\frac{l}{v_0}£©^2}$
½âµÃ£º${U}_{m}=\frac{m}{q}£¨\frac{d{v}_{0}}{l}£©^{2}$=100V£»
ËùÒÔ£¬µ±U_MN£¾U_mʱ£¬Á£×Ó´òÔÚϼ«°åÉÏ£¬²»ÄÜÉä³ö£»ËùÒÔ£¬µ±U_MN=100Vʱ£¬Á£×Ó´Ó½ðÊô°åµÄϼ«°å±ßÔµÉä³ö£¬ÕâʱÁ£×ÓËÙ¶È×î´ó£®
ÔòÁ£×ÓÉä³öʱµÄÊúÖ±·½Ïò·ÖËÙ¶È${v}_{y}=at=\frac{q{U}_{m}}{md}¡Á\frac{l}{{v}_{0}}=1¡Á1{0}^{5}m/s$£¬
ËùÒÔ£¬´øµçÁ£×ÓÉä³öµç³¡Ê±µÄ×î´óËÙ¶È${v}_{max}=\sqrt{{{v}_{y}}^{2}+{{v}_{0}}^{2}}=\sqrt{2}¡Á1{0}^{5}m/s$£»
£¨2£©Á£×ÓÔڴų¡ÖÐËùÊÜÂåÂ××ÈÁ¦×÷ΪÏòÐÄÁ¦£¬ËùÒÔ£¬$Bvq=\frac{m{v}^{2}}{r}$£¬ËùÒÔ£¬$r=\frac{mv}{Bq}$£»
ÈçͼËùʾ£¬Á£×ÓÒÔ×îСËٶȽøÈë´Å³¡ºóÏòÉÏƫת£¬ÔÈËÙÔ²ÖÜÔ˶¯µÄ°ë¾¶Îª£º${r_{min}}=\frac{{m{v_{min}}}}{qB}=\frac{{m{v_0}}}{qB}=\frac{{1¡Á{{10}^5}}}{{1¡Á{{10}^8}¡Á5¡Á{{10}^{-3}}}}m=0.2m$£»
ÓÉÓÚr_min=d£¬Òò´ËÁ£×Ó´òÔÚÆÁÉϵÄ×î¸ßµãÓëO'µãµÄ¾àÀëΪ0.2m£»
µ±Á£×Ó½øÈë´Å³¡µÄËÙ¶ÈÔö´óʱ£¬Á£×Ó´òÔÚÆÁÉϵĵãÏÂÒÆ£»
µ±Á£×ÓÒÔ×î´óËٶȽøÈë´Å³¡ºóÏòÏÂƫת£¬ÔÈËÙÔ²ÖÜÔ˶¯µÄ°ë¾¶Îª${r_{max}}=\frac{{m{v_{max}}}}{qB}=\frac{{\sqrt{2}m{v_0}}}{qB}=\sqrt{2}d$
ÒòΪv_y=v₀£¬¸ù¾Ý°ë¾¶ÓëËٶȷ½Ïò´¹Ö±£¬Óɼ¸ºÎ¹Øϵ¿ÉµÃÔ²ÖÜÔ˶¯µÄÔ²ÐÄÂäÔÚÆÁÄ»ÉϵÄQµã£¬QµãÓëM°åÔÚͬһˮƽÏßÉÏ£»
Òò´ËÁ£×Ó´òÔÚÆÁÉϵÄ×îµÍµãÓëO'µãµÄ¾àÀëΪ£º${r_{max}}-\frac{d}{2}=£¨{\sqrt{2}-\frac{1}{2}}£©d=0.18m$
×ÛÉÏ¿ÉÖª£¬´øµçÁ£×Ó´òÔÚÆÁÄ»O'µãÉÏ·½0.2mµ½O'µãÏ·½0.18mµÄ·¶Î§ÄÚ£®
´ð£º£¨1£©´øµçÁ£×ÓÉä³öµç³¡Ê±µÄ×îСËÙ¶ÈΪ1¡Á10⁵m/s£¬×î´óËÙ¶ÈΪ$\sqrt{2}¡Á1{0}^{5}m/s$£»
£¨2£©´øµçÁ£×Ó´òÔÚÆÁÄ»Éϵķ¶Î§ÎªÆÁÄ»O'µãÉÏ·½0.2mµ½O'µãÏ·½0.18mµÄ·¶Î§ÄÚ£®

µãÆÀ ±¾ÌâÒ×´íµãΪÇó×î´óËÙ¶Èʱֱ½Ó´úÈëµçѹ²îµÄ×î´óÖµ200VÇó½â£¬Òò´Ë£¬ÔÙÇóÓëÔ˶¯Ïà¹ØµÄÎÊÌâʱ£¬ÎÒÃÇÐèÒªÏÈÑéÖ¤ÎïÌåÄÜ×öÕâÖÖÔ˶¯µÄÌõ¼þÔÙÇó½â£®