求助。大学物理。有关微积分在磁场中的使用。三、 哗奥一Z成尔定T应用实例-|||-1. AL tire n与必的形场-|||-α2-|||-出 N l square =aot(pi -a) .=dfrac ({mu )_(0)}(4pi )dfrac ({d)_(0)sin alpha }({r)^2}-|||-Ⅱ =-acos a-|||-dī α dfrac (dB)(da)=dfrac (a)({sin )^2a} =int dsqrt (B)=dfrac ({a)_(0)I}(4pi )(int )_(1)^(a_{1)}dfrac (asin alpha /{sin )^2alpha }({a)^3/(sin )^2alpha }dalpha .-|||-ū =dfrac ({mu )_(0)I}(4pi a)(int )_({a)_(1)}^(a_{8)}sin ada a,=-|||-a-|||-0 -8 square .=dfrac ({H)_(0)I}(4pi a)(cos (a)_(1)-cos (a)_(2))-|||-α1 sin(π=a) .无限长 . =dfrac ({R)_(0)I}(2pi a)(a)_(1)=0-|||-T=-|||-进M .=dfrac (a)(sin a) _(1)=pi -|||-元限长直电流的磁场 =B=dfrac ({V)_(0)I}(2pi a) 公式的应用三、 哗奥一Z成尔定T应用实例-|||-1. AL tire n与必的形场-|||-α2-|||-出 N l square =aot(pi -a) .=dfrac ({mu )_(0)}(4pi )dfrac ({d)_(0)sin alpha }({r)^2}-|||-Ⅱ =-acos a-|||-dī α dfrac (dB)(da)=dfrac (a)({sin )^2a} =int dsqrt (B)=dfrac ({a)_(0)I}(4pi )(int )_(1)^(a_{1)}dfrac (asin alpha /{sin )^2alpha }({a)^3/(sin )^2alpha }dalpha .-|||-ū =dfrac ({mu )_(0)I}(4pi a)(int )_({a)_(1)}^(a_{8)}sin ada a,=-|||-a-|||-0 -8 square .=dfrac ({H)_(0)I}(4pi a)(cos (a)_(1)-cos (a)_(2))-|||-α1 sin(π=a) .无限长 . =dfrac ({R)_(0)I}(2pi a)(a)_(1)=0-|||-T=-|||-进M .=dfrac (a)(sin a) _(1)=pi -|||-元限长直电流的磁场 =B=dfrac ({V)_(0)I}(2pi a) 公式的应用三、 哗奥一Z成尔定T应用实例-|||-1. AL tire n与必的形场-|||-α2-|||-出 N l square =aot(pi -a) .=dfrac ({mu )_(0)}(4pi )dfrac ({d)_(0)sin alpha }({r)^2}-|||-Ⅱ =-acos a-|||-dī α dfrac (dB)(da)=dfrac (a)({sin )^2a} =int dsqrt (B)=dfrac ({a)_(0)I}(4pi )(int )_(1)^(a_{1)}dfrac (asin alpha /{sin )^2alpha }({a)^3/(sin )^2alpha }dalpha .-|||-ū =dfrac ({mu )_(0)I}(4pi a)(int )_({a)_(1)}^(a_{8)}sin ada a,=-|||-a-|||-0 -8 square .=dfrac ({H)_(0)I}(4pi a)(cos (a)_(1)-cos (a)_(2))-|||-α1 sin(π=a) .无限长 . =dfrac ({R)_(0)I}(2pi a)(a)_(1)=0-|||-T=-|||-进M .=dfrac (a)(sin a) _(1)=pi -|||-元限长直电流的磁场 =B=dfrac ({V)_(0)I}(2pi a) 公式的应用三、 哗奥一Z成尔定T应用实例-|||-1. AL tire n与必的形场-|||-α2-|||-出 N l square =aot(pi -a) .=dfrac ({mu )_(0)}(4pi )dfrac ({d)_(0)sin alpha }({r)^2}-|||-Ⅱ =-acos a-|||-dī α dfrac (dB)(da)=dfrac (a)({sin )^2a} =int dsqrt (B)=dfrac ({a)_(0)I}(4pi )(int )_(1)^(a_{1)}dfrac (asin alpha /{sin )^2alpha }({a)^3/(sin )^2alpha }dalpha .-|||-ū =dfrac ({mu )_(0)I}(4pi a)(int )_({a)_(1)}^(a_{8)}sin ada a,=-|||-a-|||-0 -8 square .=dfrac ({H)_(0)I}(4pi a)(cos (a)_(1)-cos (a)_(2))-|||-α1 sin(π=a) .无限长 . =dfrac ({R)_(0)I}(2pi a)(a)_(1)=0-|||-T=-|||-进M .=dfrac (a)(sin a) _(1)=pi -|||-元限长直电流的磁场 =B=dfrac ({V)_(0)I}(2pi a) 公式的应用三、 哗奥一Z成尔定T应用实例-|||-1. AL tire n与必的形场-|||-α2-|||-出 N l square =aot(pi -a) .=dfrac ({mu )_(0)}(4pi )dfrac ({d)_(0)sin alpha }({r)^2}-|||-Ⅱ =-acos a-|||-dī α dfrac (dB)(da)=dfrac (a)({sin )^2a} =int dsqrt (B)=dfrac ({a)_(0)I}(4pi )(int )_(1)^(a_{1)}dfrac (asin alpha /{sin )^2alpha }({a)^3/(sin )^2alpha }dalpha .-|||-ū =dfrac ({mu )_(0)I}(4pi a)(int )_({a)_(1)}^(a_{8)}sin ada a,=-|||-a-|||-0 -8 square .=dfrac ({H)_(0)I}(4pi a)(cos (a)_(1)-cos (a)_(2))-|||-α1 sin(π=a) .无限长 . =dfrac ({R)_(0)I}(2pi a)(a)_(1)=0-|||-T=-|||-进M .=dfrac (a)(sin a) _(1)=pi -|||-元限长直电流的磁场 =B=dfrac ({V)_(0)I}(2pi a) 公式的应用三、 哗奥一Z成尔定T应用实例-|||-1. AL tire n与必的形场-|||-α2-|||-出 N l square =aot(pi -a) .=dfrac ({mu )_(0)}(4pi )dfrac ({d)_(0)sin alpha }({r)^2}-|||-Ⅱ =-acos a-|||-dī α dfrac (dB)(da)=dfrac (a)({sin )^2a} =int dsqrt (B)=dfrac ({a)_(0)I}(4pi )(int )_(1)^(a_{1)}dfrac (asin alpha /{sin )^2alpha }({a)^3/(sin )^2alpha }dalpha .-|||-ū =dfrac ({mu )_(0)I}(4pi a)(int )_({a)_(1)}^(a_{8)}sin ada a,=-|||-a-|||-0 -8 square .=dfrac ({H)_(0)I}(4pi a)(cos (a)_(1)-cos (a)_(2))-|||-α1 sin(π=a) .无限长 . =dfrac ({R)_(0)I}(2pi a)(a)_(1)=0-|||-T=-|||-进M .=dfrac (a)(sin a) _(1)=pi -|||-元限长直电流的磁场 =B=dfrac ({V)_(0)I}(2pi a) 公式的应用你好。微积分在大学物理电磁学中的使用。第一张，推出载流直导线的磁场，我可以理解。在后面的题目中，或是求一个面，或是求一个圆圈，请问要怎样使用定积分去推导出来？https:/zhidao.baidu.com/question/1382000517512775940.htmlhttps:/zhidao.baidu.com/question/1365902875976612019.html