Please explain, Thanks A charged particle (q = +3.1 mu C) moves at speed v0 = 49

Question

Please explain, Thanks

A charged particle (q = +3.1 mu C) moves at speed v0 = 49 m/s in the +x-direction. At x = 0 it enters a region where a constant magnetic field B = 3.2 T which is directed in the +z direction as shown in the diagram below. (The y-axis points into the page.) What is the magnitude of the magnetic force on the particle? Now a uniform electric field is applied so that the net force on this particle is zero. At the moment the particle first enters the region of the magnetic field, what direction is the force due to the magnetic field on the particle? Direction of force due to magnetic field = What are the components of the electric field?

Explanation / Answer

The magnitude of magnetic force on the particle is F =qvB =(3.1*10-6)(49)(3.2) =4.86*10-4N

The direction force due to magnetic field will be along the y -direction from flemings right hand rule

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