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Prep101comsolutions 33 prep101 phys1026 april exam

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Unformatted text preview: ns will be posted at www.prep101.com/solutions 31 ©Prep101 Phys1026 April Exam Solutions Question 8: Solution: Due to the different directions of current in the loop, one side of the loop is attracted to the large wire carrying wire, while the other repulsed. Question 9: Solution: In this figure, the current in the two wires are in opposite directions. To suspend the top wire, the magnetic force per unit length must be equal to the wire’s weight per unit length. Furthermore, we must orient the top wire’s current into the page, in order to get the force pointing upwards. The magnetic force exerted on a conductor of length l carrying current It perpendicular to a field of size Bl is F = BlItl. Therefore, the magnetic force per unit length acting on the top wire is: Solutions will be posted at www.prep101.com/solutions 32 ©Prep101 Phys1026 April Exam Solutions Question 10: Solution: First we find the speed of the proton from its kinetic energy. It is easier to calculate b.) before a.), therefore we need to know the orbital radius of the proton in the B – field. B – field directed out of page (y or k axis) As the proton passes through the B – field, the central force is balanced by the magnetic force, and as the velocity is normal to this field: Therefore, from the path length over which this proton is exposed to the magnetic field along with the orbital radius, we are able to determine the angle between the initial and final velocity vectors. It is now straightforward to calculate the x-axis component of the protons’ momentum: 9.11 Answer If the electron is travelling in a circle in a magnetic field, it is undergoing cyclotron motion. This is where the magnetic field causes it to move in a circle, so the magnetic force is the centripetal force FB = FC qvB = mv2/r qB = mv/r v = rqB /m To find out how often it goes round, we look at Solutions will be posted at www.prep101.com/solutions 33 ©Prep101 Phys1026 April Exam Solutions T = distance / v = 2π r / v = 2π r / (rqB/m) = 2π m / qB So then the frequency is f = 1/T f= qB 2πm = 1.6 x 10-19 x 1 / (2 x 3.1 x 9.1 x 10-31) = 2.8 x 1010 Hz The electron completes 2.8 x 1010 orbits every second. The current through a plane is the flow of charge per second. The electron passes through A 2.8 x 1010 times a second, so the total current is I = 2.8 x 1010 x 1.6 x 10-19 = 5 x 10-9 A The electron passes through B twice an orbit, once in each direction – so the net flow of charge is zero (just as much flows to the left as to the right). The current through B is zero. 9.12 Answer e) Magnetic fields only exert forces on moving charges – the sphere is not moving, so there is no force. 9.13 Answer Strategy -develop equation for B at centre of loop -let Bwire = Bloop for a net field strength of zero -use RHR for current direction in the coil Long wire: B = μoIw/2πr and setting the angle = 2π) equating the two Loop: B = μoIL/2R ( from the equation for an arc, μoIw/2πr = μoIL/2R Iw/π(r+R) = IL/R ---solve for r r = (Iw R) / (π IL) – R Solutions will be posted at www.prep101.com/solutions 3...
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This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.

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