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A charged particle moves with velocity 5 km/s making angle 50° with the external magnetic field...

  1. A charged particle moves with velocity 5 km/s making angle 50° with the external magnetic field of magnitude of 0.6 T (see figure below). The field exerts the 3.5 N force on the charge. The direction of the force is out of the page/screen. What is the magnitude and sign of the charge? The magnetic force would be zero if: v was parallel to the field; (a) v was directed out of page;(b)v was directed into the page(c)v=0;(d) A and D;(e)B and C.
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Answer #1

F=qobsino Force on charged particle in mo nou ving magnetic field F = qb xB) 3.5 = q 15000 *0.6 ein50) q= 1.523 x 10°C if we

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