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Unformatted text preview: 47. (a) We denote the B -ﬁelds at point P on the axis due to the solenoid and the wire as Bs and Bw ,
respectively. Since Bs is along the axis of the solenoid and Bw is perpendicular to it, Bs ⊥ Bw ,
respectively. For the net ﬁeld B to be at 45◦ with the axis we then must have Bs = Bw . Thus,
Bs = µ0 is n = Bw = µ0 iw
2πd which gives the separation d to point P on the axis:
= 4.77 cm .
2π (20.0 × 10−3 A)(10 turns/cm) (b) The magnetic ﬁeld strength is
2(4π × 10−7 T · m/A)(20.0 × 10−3 A)(10 turns/0.0100 m)
= 3.55 × 10−5 T . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
- Fall '08