hw14 - Physics 21 Fall, 2008 Solution to HW-14 B of Finite...

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Unformatted text preview: Physics 21 Fall, 2008 Solution to HW-14 B of Finite Wire A steady current I is flowing through a straight wire of finite length. (a) Find B 1 , the magnetic field generated by this wire at a point P located a distance x from the center of the wire. Assume that at P the angle subtended from the midpoint of the wire to each end is m as shown in part (a) of the diagram. (b) Now find B 2 , the magnetic field generated by this wire at a point P located a distance x from either end of the wire. Assume that at P the angle subtended from the end of the wire to the other end is end as shown in part (b) of the diagram We can apply the method used in class by relating the angles m and end with the wire length. For panel (a), tan m = a/x a = x tan m For panel (b), tan end = a/x a = x tan end Begin with the Biot-Savart Law d B = 4 Id l ( r r ) | r r | 3 . Let the current point r be on the y axis as y j ; the field point r is at P or x i , and d l = dy j : d B = 4 Idy j ( x i y j ) ( x 2 + y 2 ) 3 / 2 = I 4 x dy ( x 2 + y 2 ) 3 / 2 k We can use this expression for part (a) and for part (b): (a) Integrating from a to + a gives B = I 4 x Z + a a dy ( x 2 + y 2 ) 3 / 2 k = I 2 a x p x 2 + y 2 k Substituting a = x tan m gives B = I 2 tan...
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hw14 - Physics 21 Fall, 2008 Solution to HW-14 B of Finite...

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