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Physics 8 Online
OpenStudy (anonymous):

An alpha particle carries a charge of 3.2 × 10^–19 C. It experiences a force of 6.5 × 10^–14 N as it moves at a right angle through a magnetic field that has a magnitude of 1.4 × 10^–3 T. How fast is the particle moving? 2.9 × 10^-35 m/s 1.5 × 10^-35 m/s 1.5 × 10^8 m/s 2.8 × 10^8 m/s

OpenStudy (anonymous):

The equation for Force on a moving charge is: \[F=qv \times B\] Since it is at a right angle(90deg) with the B (magnetic) field, the cross product can go away and you're left with \[F = qvB\] Solve for v: \[v = F/qB\] v = (6.5x10^-14)/[(3.2x10^-19)(1.4x10^-3)] v = 1.45x10^8 m/s which can be rounded up to 1.5x10^8 m/s

OpenStudy (anonymous):

thank you so much :)

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