Final No Date 2

Final No Date 2

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Unformatted text preview: 4 ©Prep101 Phys1026 April Exam Solutions = (6.0A)(0.05m) / [π (1.0A)] – (0.05m) = 0.045m r = 4.5cm Using RHR, the direction of B due to the long wire is into the page, therefore, the direction of B due to the loop must be out of the page, to satisfy the Bnet of zero, and the current should flow counter-clockwise. Solutions will be posted at www.prep101.com/solutions 35 ©Prep101 Phys1026 April Exam Solutions 10. INDUCTION AND INDUCTANCES Question 1 Solution This is a question about energy transfer and power in a changing loop. Key point: The power dissipated in the loop (in the resistor) is equal to the work done by the force Ploop = Pforce The power dissipated in the loop can be found from ε= dΦ d ( BA) ds(t) = = Ba = Bav dt dt dt ∴I = and ε = IR Bav R ∴ P = vI = I 2 R = B2a 2 v 2 R so Ploop= PForce B2a2v2/R = Fv F = B2a2v/R Solutions will be posted at www.prep101.com/solutions 36 ©Prep101 Phys1026 April Exam Solutions Question 2 Answer a) As it enters the field Call x the distance that the leading edge has penetrated into the strong B field region. The flux through the loop increases in time so the voltage: is induced in the coil, producing a counterclockwise currentthat creates its own B – field that opposes this change. b) The flux through the loop is constant when the loop is wholly inside the B – field region, therefore the induced emf is zero. c) The emf is the same size as in part a) but in the opposite direction, clockwise. Solutions will be posted at www.prep101.com/solutions 37 ©Prep101 Phys1026 April Exam Solutions Question 3 Solution a) The current will be moving down the metal bar. Use the right-hand rule; your fingers point in the direction of the magnetic field, into the page. Your thumb points in the direction of the current along the bar, which is down. Your palm is facing to the right which is the direction of the magnetic force which causes the bar to move. The bar moves to the right. b) The bar moves to the right and this increases the flux going into the page. The magnetic field generated by the induced current needs to oppose this change in flux, so the magnetic field generated by the induced current needs to point out of the page. By the other right-hand rule, this means the current is going counter-clockwise around the closed circuit. Solutions will be posted at www.prep101.com/solutions 38 ©Prep101 Phys1026 April Exam Solutions Question 4 Solution a) The induced emf is given by ε= − dΦ B dt The magnetic flux is ΦB = BA and the only a...
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This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.

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