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Unformatted text preview: 5 The magnitude of the magnetic field inside the solenoid is B = 0 nis , where n is the number of turns per unit length and is is the current. The field is parallel to the solenoid axis, so the 2 2 flux through a cross section of the solenoid is B = As B = 0 rs nis , where As (= rs ) is the cross-sectional area of the solenoid. Since the magnetic field is zero outside the solenoid, this is also the flux through the coil. The emf in the coil has magnitude E= and the current in the coil is N dB dis 2 = 0 rs N n dt dt 2 E 0 rs N n dis = , R R dt where N is the number of turns in the coil and R is the resistance of the coil. The current changes linearly by 3.0 A in 50 ms, so dis /dt = (3.0 A)/(50 10-3 s) = 60 A/s. Thus ic = ic = (4 10-7 T m/A)(0.016 m)2 (120)(220 102 m-1 ) (60 A/s) = 3.0 10-2 A . 5.3 ...
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This lab report was uploaded on 04/15/2008 for the course PHYS 101 taught by Professor Reich during the Spring '08 term at Johns Hopkins.
- Spring '08