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Unformatted text preview: ni sol = 6 . 00 A 2 (10 . 10 2 m 1 )(20 . 10 3 A) = 4 . 77 10 2 m = 4 . 77 cm . This distance is less than the radius of the solenoid, so the point is indeed inside as we assumed. (b) The magnitude of the either field at the point is B sol = B wire = ni sol = (4 10 7 T m / A)(10 . 10 2 m 1 )(20 . 10 3 A) = 2 . 51 10 5 T . Each of the two fields is a vector component of the net field, so the magnitude of the net field is the square root of the sum of the squares of the individual fields: B = 2(2 . 51 10 5 T) 2 = 3 . 55 10 5 T....
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 Spring '08
 Reich

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