ch29-p053 - ni sol = 6 . 00 A 2 (10 . 10 2 m 1 )(20 . 10 3...

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53 (a) Assume that the point is inside the solenoid. The field of the solenoid at the point is parallel to the solenoid axis and the field of the wire is perpendicular to the solenoid axis. The net field makes an angle of 45 with the axis if these two fields have equal magnitudes. The magnitude of the magnetic field produced by a solenoid at a point inside is given by B sol = µ 0 i sol n , where n is the number of turns per unit length and i sol is the current in the solenoid. The magnitude of the magnetic field produced by a long straight wire at a point a distance r away is given by B wire = µ 0 i wire / 2 πr , where i wire is the current in the wire. We want µ 0 ni sol = µ 0 i wire / 2 πr . The solution for r is r = i wire 2
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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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