29. Find the momentum of a charge of (3.2 X 10-19) that moves perpendicular to a

Question

29. Find the momentum of a charge of (3.2 X 10-19) that moves perpendicular to a uniform magnetic field B of magnitude 0.20 T and has radius of curvature of 4.0 cm. Remember, momentum = mv



30. An electron of energy 400 eV moves at right angles to a uniform magnetic field along an arc of radius 2.0 cm. A second electron moves at right angles to the field along an arc of radius 3.2 cm. What is the energy of the second electron?



32. The magnetic field at the center of a 200 turn coil of radius 12 cm is 6.6 mT. Find the current in the coil.

Explanation / Answer

29. momentum = mv = qBr = 3.2*10^-19*.2*.04 = 2.56*10^-21 30. Energy = (3.2/2)^2*400 = 1024 eV 32. current = 6.6*10^-3*2*.12/(200*1.26*10^-6) = 6.29 A

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