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phys1002_sln05_sm2_2007

phys1002_sln05_sm2_2007 - PHYS1002 TUTORIAL 5 SOLUTIONS...

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PHYS1002 TUTORIAL 5 SOLUTIONS Semester 2, 2007 1. MAGNETIC MATERIALS We have B τ μ = × arrowrightnosp arrowrightnosp arrowrightnosp where μ arrowrightnosp is the magnetic moment of the magnet, B arrowrightnosp is the magnetic field of the solenoid, and τ arrowrightnosp is the torque experienced by the magnet. Here the magnetic moment will be along the length of the magnet and hence perpendicular to B arrowrightnosp . So τ μ = B and therefore / B μ τ = . The magnetic field of the solenoid is given by B n I o = μ 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 π . Hence μ = × × = 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 = = × × × = × μ π 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. FARADAY’S LAW I Let ˆ n be a unit vector perpendicular to the plane of the loop. At time t = 0, we choose ˆ n to point in the positive z direction. Let θ be the angle between B arrowrightnosp and ˆ n . Since the loop is set to rotate at a constant angular frequency ω and since 0 θ = at t = 0, it follows that t θ ω = for 0 t . The magnetic flux through the loop is then 2 ˆ cos cos( ) S B ndA BA B r t φ θ π ω = = =
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