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F10HW09 - 1 Problem 5.13 trting with eqution SFQP in tksonX...

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1 Problem 5.13 Starting with equation 5.32 in Jackson: ~ A ( ~ r ) = 0 4 Z V ~ j ( ~ r ) 1 j ~ r   ~ r 0 j d 3 r 0 = 0 4 Z 2 ' 0 =0 Z 1 cos 0 =   1 0 @ ~! ¢ ~ r 0 | {z } !r 0 sin 0 ^ ' 0 1 A 1 X ` =0 ` X m =   ` 4 2 ` + 1 r ` < r ` +1 > Y m £ ` ( 0 ; ' 0 ) Y m ` ( ; ' ) ! r 0 2 d (cos 0 ) d' 0 r 0 = a = 0 4 !a 3 1 X ` =0 ` X m =   ` 4 2 ` + 1 r ` < r ` +1 > Z 2 ' 0 =0 Z 1 cos 0 =   1 Y m £ ` ( 0 ; ' 0 ) Y m ` ( ; ' ) sin 0 ^ ' 0 d (cos 0 ) d' 0 = 0 !a 3 1 X ` =0 ` X m =   ` 1 2 ` + 1 r ` < r ` +1 > Y `;m ( ; ' ) Z 2 ' 0 =0 Z 1 cos 0 =   1 Y m £ ` ( 0 ; ' 0 ) sin 0 ^ ' 0 d (cos 0 ) d' 0 (1) Since ^ ' 0 =   sin ' 0 ^ x + cos ' 0 ^ y , the integral becomes: Z 2 ' 0 =0 Z 1 cos 0 =   1 Y m £ ` ( 0 ; ' 0 ) (   sin 0 sin ' 0 ^ x + sin 0 cos ' 0 ^ y ) d (cos 0 ) d' 0 We know that sin 0 sin ' 0 = sin e i'   e   i' 2 i = 1 i q 2 3 ¢   Y 1 1 ( 0 ; ' 0 )   Y   1 1 ( 0 ; ' 0 ) £ and sin 0 cos ' 0 = sin e i' + e   i' 2 = q 2 3 ¢   Y 1 1 ( 0 ; ' 0 ) + Y   1 1 ( 0 ; ' 0 ) £ . By exploiting orthogonality, the integral becomes: r 2 3 Z 2 ' 0 =0 Z 1 cos 0 =   1 Y m £ ` ( 0 ; ' 0 ) ¢   i   Y l 1 ( 0 ; ' 0 ) + Y   1 1 ( 0 ; ' 0 ) ¡ ^ x +     Y l 1 ( 0 ; ' 0 ) + Y   1 1 ( 0 ; ' 0 ) ¡ ^ y £ d (cos 0 ) d' 0 = r 2 3 ¢   i   1 1 +   1 1 ¡ ^ x +     1 1 +   1 1 ¡ ^ y £ Plugging this into equation (1) yields: ~ A ( ~ r ) = 0 !a 3 1 2(1) + 1 r 1 < r (1)+1 > r 2 3 ¢   i   Y l 1 ( ; ' ) + Y   1 1 ( ; ' ) ¡ ^ x +     Y l 1 ( ; ' ) + Y   1 1 ( ; ' ) ¡ ^ y £ ~ A ( ~ r ) = 0 !a 3 1 2(1) + 1 r 1 < r (1)+1 > [   sin sin ' ^ x + sin cos ' ^ y ] ~ A ( ~ r ) = 0 !a 3 1 3 r < r 2 > sin [   sin ' ^ x + cos ' ^ y ] | {z } ^ ' ~ A ( ~ r ) = 0 !a 3 1 3 r < r 2 > sin ^ ' 1
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Inside the sphere: ~ A in ( ~ r ) = 1 3 0 !a 3 r a 2 sin ^ ' = 1 3 0 !ar sin ^ ' Outside the sphere: ~ A out ( ~ r ) = 1 3 0 !a 3 a r 2 sin ^ ' = 1 3 0 ! a 4 r 2 sin ^ ' ~ B = r ¢ ~ A = 1 r sin @ @ (sin A ' )^ r   1 r @ @r ( rA ' ) ^ ~ B in = 1 r sin @ 1 3 0 !ar sin 2 ^ r   1 r @ 1 3 0 !ar 2 sin ^ = 1 r sin 1 3 0 !ar 2 sin cos ^ r   1 r 1 3 0 !a 2 r sin ^ = 2 3 0 !a cos ^ r   sin ^ ~ B out = 1 r sin @ 1 3 0 ! a 4 r 2 sin 2 ^ r   1 r @ 1 3 0 ! a 4 r sin ^ = 1 r sin 1 3 0 ! a 4 r 2 2 sin cos ^ r   1 r 1 3 0 !   a 4 r 2 sin ^ = 1 3 0 ! a 4 r 3 2 cos ^ r + sin ^ 2
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2 Problem 5.15 2.1 First, we use Amp ere's Law to determine ~ B for a single wire carrying current I ^ z : I ~ B ¡ d ~ l = 0 I end B 2 r = 0 I = ) ~ B = 0 I 2 r ^ ' where the direction of ~ B comes from applying the right-hand rule. Since ~ H = ~ B= 0 : ~ H = I 2 r ^ ' The magnetic scalar potential ¨
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F10HW09 - 1 Problem 5.13 trting with eqution SFQP in tksonX...

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