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Transcript

PHYS1022 Electricity and Magnetism Problem Sheet 7 - tutorials 1. At any point in space, the electric field E is defined to be in the direction of the electric force on a positively charged particle at that point. Why not define the magnetic field B to be in the direction of the magnetic force F on a moving, positively charged particle? 2. A charged particle moves through a region of space with constant velocity (magnitude and direction). If the external magnetic field is zero in this region, can you conclude that the external electric field in the region is also zero? Explain. If the external electric field is zero in the region, can you conclude that the external magnetic field in the region is also zero? 3. An electron from the sun with a speed of 1 × 107 m/s enters the earth’s magnetic field high above the equator where the magnetic field is 4 × 10−7 T. The electron moves nearly in a circle, except for a small drift along the direction of the earth’s magnetic field that will take the electron towards the north pole. What is the radius of the circular motion? Derive the equation that you use. 4. A beam of protons moves along the x-axis in the positive x direction with a speed of 12.4 km/s through a region of crossed fields balanced for zero deflection. (a) If there is a magnetic field of magnitude 0.85 T in the positive y direction, find the magnitude and direction of the electric field. (b) Would electrons of the same velocity be deflected by these fields? If so, in what direction? 1 Solutions 1. The force is given by the cross product of velocity and magnetic field (F = q v x B) – so the direction of the force depends on the direction of velocity and we can not use a notation like that for electric field. 2. An electric field would exert a force on the charge which would need to be balanced by another force to have no acceleration. Provided there are no other forces (such as gravity) acting then E=0. If E=0 (and no other forces but B act) there can still be a B field in the direction of v (so the cross product is zero). 3. For circular motion in a magnetic field Newton 2 gives m v2/r = q v B so r = m v/ q B r = (9.11x10-31kg x 107ms-1) / (1.6x10-19C x 4x10-7T) = 142m 4. The net force must be zero so that qvB=qE E = 0.85T x 12.4x103ms-1 = 10.5 kVm-1 Note the magnetic force is in the +ve z direction… so the E field must point in the –ve z-direction. 2