# hw13 - Physics 21 Fall 2008 Solution to HW-13 28-11 A long...

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Unformatted text preview: Physics 21 Fall, 2008 Solution to HW-13 28-11 A long, straight wire lies along the z axis and carries a 4.10 A current in the + z direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.400 mm segment of the wire centered at the origin: (a) x = 2 . 00 m, y = 0, z = 0, (b) x = 0, y = 2 . 00 m, z = 0, (c) x = 2 . 00 m, y = 2 . 00 m, z = 0, (d) x = 0, y = 0, z = 2 . 00 m. The length (0.400 mm) of the wire segment with the cur- rent is tiny compared to the distance (two or more meters) from the segment to the points where the field is wanted, so we can use the differential form of the Biot-Savart Law: d B = µ 4 π Id l × ( r- r ) | r- r | 3 . The current point r is at the origin, so r = 0. Also, d l = dl ˆ k , and dl = 4 . 00 × 10 − 4 m. Writing all distances in meters, and using ˆ k × ˆ i = ˆ j , ˆ k × ˆ j =- ˆ i , ˆ k × ˆ k = 0 , we have: (a) for the field point r = 2 . 00 ˆ i , d B = µ 4 π Idl ˆ k × 2 ˆ i 2 3 = 4 . 10 × 10 − 11 T ˆ j (b) for the field point r = 2 . 00 ˆ j , d B = µ 4 π Idl ˆ k × 2 ˆ j 2 3 =- 4 . 10 × 10 − 11 T ˆ i (c) for the field point r = 2 . 00 ˆ i + 2 . 00 ˆ j , d B = µ 4 π Idl ˆ k × (2 ˆ i + 2 ˆ j ) (4 + 4) 3 / 2 = 1 . 45 × 10 − 11 T(- ˆ i + ˆ j ) (d) for the field point r = 2 . 00 ˆ k , d B = µ 4 π Idl ˆ k × 2 ˆ k 2 3 = 0 28-19 A long, straight wire lies along the y axis and car-...
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hw13 - Physics 21 Fall 2008 Solution to HW-13 28-11 A long...

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