phys1002_sln05_sm2_2007

phys1002_sln05_sm2_2007 - PHYS1002 TUTORIAL 5 SOLUTIONS...

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Unformatted text preview: PHYS1002 TUTORIAL 5 SOLUTIONS Semester 2, 2007 1. MAGNETIC MATERIALS We have B = a a a where a is the magnetic moment of the magnet, B a is the magnetic field of the solenoid, and a is the torque experienced by the magnet. Here the magnetic moment will be along the length of the magnet and hence perpendicular to B a . So = B and therefore / B = . The magnetic field of the solenoid is given by B n I o = where n is the number of turns per meter and I is the current through the solenoid. Here B T = = 4 10 1000 10 1 26 10 7 2 . Hence = = 18 10 1 26 10 1 43 2 2 2 . / . . A m The magnetisation is the magnetic moment per unit volume, M d dV A m = = = 1 43 0 1 5 10 7 3 10 3 2 5 . . . / e j Since M di d = where i is the Amperean current and is the length of the magnet, we have i A = = 7 3 10 0 1 7 3 10 5 4 . . . . 2. FARADAYS LAW I Let n be a unit vector perpendicular to the plane of the loop. At time t = 0, we choose n to point in the positive z direction. Let be the angle between B a and n . Since the loop is set to rotate at a constant angular frequency and since = at t = 0, it follows that t = for t . The magnetic flux through the loop is then 2 cos cos( ) S B ndA BA B r t =...
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phys1002_sln05_sm2_2007 - PHYS1002 TUTORIAL 5 SOLUTIONS...

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