# 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 θ 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 θ end as shown in part (b) of the diagram We can apply the method used in class by relating the angles θ m and θ end with the wire length. For panel (a), tan θ m = a/x ⇒ a = x tan θ m For panel (b), tan θ end = a/x ⇒ a = x tan θ end Begin with the Biot-Savart Law d B = µ 4 π Id l × ( r − r ) | r − r | 3 . Let the current point r be on the y axis as y ˆ j ; the field point r is at P or x ˆ i , and d l = dy ˆ j : d B = µ 4 π Idy ˆ j × ( x ˆ i − y ˆ j ) ( x 2 + y 2 ) 3 / 2 = − µ I 4 π x dy ( x 2 + y 2 ) 3 / 2 ˆ k We can use this expression for part (a) and for part (b): (a) Integrating from − a to + a gives B = − µ I 4 π x Z + a − a dy ( x 2 + y 2 ) 3 / 2 ˆ k = − µ I 2 π a x p x 2 + y 2 ˆ k Substituting a = x tan θ m gives B = − µ I 2 π tan θ...
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hw14 - Physics 21 Fall 2008 Solution to HW-14 B of Finite...

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