Problem: I 1 = 2 A ; I 2 = 9 A ; P 1 = ( 40 , 60 ) c m ; P 2 = ( 40 , − 30 ) c m ; H ⃗ ( 25 , 0 ) = ? I_1=2A; I_2=9A; P_1=(40,60)cm; P_2=(40,-30)cm;\vec H(25,0)=? I 1 = 2 A ; I 2 = 9 A ; P 1 = ( 40 , 60 ) c m ; P 2 = ( 40 , − 30 ) c m ; H ( 25 , 0 ) = ?
Solution: For an infinitely long conductor the magnetic field at any point r r r ( 1 ) H ⃗ = 1 4 π 2 I ⋅ l ^ × r ^ r (1) \vec H=\frac{1}{4\pi}\frac{2 I \cdot \hat l ×\hat r}{r} ( 1 ) H = 4 π 1 r 2 I ⋅ l ^ × r ^ l ^ \hat l l ^ r ^ \hat r r ^ l ^ = − k ^ \hat l=-\hat k l ^ = − k ^ r ⃗ 1 = 25 i ^ − 40 i ^ − 60 j ^ = − 15 i ^ − 60 j ^ \vec r_1=25 \hat i -40\hat i - 60 \hat j=-15 \hat i-60\hat j r 1 = 25 i ^ − 40 i ^ − 60 j ^ = − 15 i ^ − 60 j ^
(2) r 1 = 1 5 2 + 6 0 2 c m = 61.8 c m = 0.618 m ; r ^ 1 = r ⃗ 1 r 1 = − 15 i ^ + 60 j ^ 61.8 = − 0.24 i ^ − 0.97 j ^ r_1=\sqrt{15^2+60^2} cm=61.8 cm=0.618 m; \hat r_1=\frac {\vec r_1}{r_1}=-\frac{15 \hat i +60 \hat j}{61.8}=-0.24 \hat i-0.97 \hat j r 1 = 1 5 2 + 6 0 2 c m = 61.8 c m = 0.618 m ; r ^ 1 = r 1 r 1 = − 61.8 15 i ^ + 60 j ^ = − 0.24 i ^ − 0.97 j ^
Vector product can be calculated using the rule k ^ × i ^ = j ^ \hat k ×\hat i=\hat j k ^ × i ^ = j ^ k ^ × j ^ = − i ^ \hat k ×\hat j=-\hat i k ^ × j ^ = − i ^
(3) H ⃗ 1 = 1 4 π 2 I 1 ( − k ^ ) × ( − 0.24 i ^ − 0.97 j ^ ) r 1 = 1 4 π 4 A 0.618 m ⋅ ( − 0.97 i ^ + 0.24 j ^ ) = = 0.515 ⋅ ( − 0.97 i ^ + 0.24 j ^ ) A m − 1 = ( − 0.500 i ^ + 0 , 124 j ^ ) A m − 1 \vec H_1=\frac{1}{4\pi}\frac{2I_1 (-\hat k) ×(-0.24 \hat i-0.97 \hat j)}{r_1}=\frac{1}{4\pi}\frac{4A }{0.618 m}\cdot (-0.97 \hat i+0.24\hat j)=\\=0.515\cdot (-0.97 \hat i+0.24\hat j)Am^{-1}=(-0.500\hat i +0,124 \hat j )Am^{-1} H 1 = 4 π 1 r 1 2 I 1 ( − k ^ ) × ( − 0.24 i ^ − 0.97 j ^ ) = 4 π 1 0.618 m 4 A ⋅ ( − 0.97 i ^ + 0.24 j ^ ) = = 0.515 ⋅ ( − 0.97 i ^ + 0.24 j ^ ) A m − 1 = ( − 0.500 i ^ + 0 , 124 j ^ ) A m − 1
For second conductor we have
r ⃗ 2 = 25 i ^ − 40 i ^ + 30 j ^ = − 15 i ^ + 30 j ^ ; r 2 = 1 5 2 + 3 0 2 = 33.5 c m = 0.335 m \vec r_2=25\hat i -40 \hat i + 30\hat j=-15 \hat i +30 \hat j; r_2=\sqrt{15^2+30^2}=33.5 cm=0.335m r 2 = 25 i ^ − 40 i ^ + 30 j ^ = − 15 i ^ + 30 j ^ ; r 2 = 1 5 2 + 3 0 2 = 33.5 c m = 0.335 m
r ^ 2 = r ⃗ 2 r 2 = − 15 i ^ + 30 j ^ 33.5 = − 0.447 i ^ + 0.894 j ^ \hat r_2=\frac{\vec r_2}{r_2}=\frac{-15 \hat i +30 \hat j}{33.5}=-0.447 \hat i+0.894 \hat j r ^ 2 = r 2 r 2 = 33.5 − 15 i ^ + 30 j ^ = − 0.447 i ^ + 0.894 j ^
(4) H ⃗ 2 = 1 4 π 2 I 2 ( − k ^ ) × ( − 0.447 i ^ + 0.894 j ^ ) r 2 = 1 4 π 18 A 0.335 m ⋅ ( 0.894 i ^ + 0.447 j ^ ) = = 4.28 ⋅ ( 0.894 i ^ + 0.447 j ^ ) A m − 1 = ( 3.82 i ^ + 1.91 j ^ ) A m − 1 \vec H_2=\frac{1}{4\pi}\frac{2I_2 (-\hat k) ×(-0.447 \hat i+0.894 \hat j)}{r_2}=\frac{1}{4\pi}\frac{18A }{0.335 m}\cdot (0.894 \hat i+0.447\hat j)=\\=4.28\cdot (0.894 \hat i+0.447\hat j)Am^{-1}=(3.82\hat i +1.91 \hat j)Am^{-1} H 2 = 4 π 1 r 2 2 I 2 ( − k ^ ) × ( − 0.447 i ^ + 0.894 j ^ ) = 4 π 1 0.335 m 18 A ⋅ ( 0.894 i ^ + 0.447 j ^ ) = = 4.28 ⋅ ( 0.894 i ^ + 0.447 j ^ ) A m − 1 = ( 3.82 i ^ + 1.91 j ^ ) A m − 1
The sketch explains the resulting picture of the fields
Now we can get overall magnetic field
H ⃗ = H ⃗ 1 + H ⃗ 2 = ( − 0.500 i ^ + 0 , 124 j ^ ) A m − 1 + ( 3.82 i ^ + 1.91 j ^ ) A m − 1 = = ( 3.32 i ^ + 2.03 j ^ ) A m − 1 \vec H=\vec H_1+\vec H_2=(-0.500\hat i +0,124 \hat j )Am^{-1}+(3.82\hat i +1.91 \hat j)Am^{-1}=\\=(3.32 \hat i+2.03 \hat j) Am^{-1} H = H 1 + H 2 = ( − 0.500 i ^ + 0 , 124 j ^ ) A m − 1 + ( 3.82 i ^ + 1.91 j ^ ) A m − 1 = = ( 3.32 i ^ + 2.03 j ^ ) A m − 1
Answer: The net magnetic field at the point (25,0) in unit vector notation is ( 3.32 i ^ + 2.03 j ^ ) A m − 1 (3.32 \hat i+2.03 \hat j) Am^{-1} ( 3.32 i ^ + 2.03 j ^ ) A m − 1
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