Moving Charges and Magnetism Test

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As a charged particle $$q$$ moving with a velocity $$\vec { v } $$ enters a uniform magnetic field $$\vec { B } $$, it experiences a force $$\vec { f } =q(\vec { v } \times \vec { B } )$$.For $$\theta={0}^{o}$$ or $${180}^{o}$$, $$\theta$$ being the angle between $$\vec { v } $$ and $$\vec { B } $$, force experienced is zero and the particle passes undeflected. For $$\theta={90}^{o}$$, the particle moves along a circular arc and the magnetic force $$(qvB)$$ provieds the necessary centripetal force $$(\cfrac{m{v}^{2}}{r})$$. For other values of $$\theta$$ ($$\theta\ne {0}^{o}, {180}^{o}, {90}^{o}$$), the charged particle moves along a helical path which is the resultant motion of simultaneous circular and translational motions. Suppose a particle, that carries a charge of magnitude $$q$$ and has a mass $$4\times {10}^{-15}kg$$, is moving in a region containing a uniform magnetic field $$\vec { B } =-0.4\hat { k } T$$. At a certain instant, velocity of the particle is $$\hat { k } =(8\hat { i } -6\hat { j } +4\hat { k } )\times {10}^{6}m/s$$ and force acting on it has a magnitude $$1.6N$$

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Directions For Questions

A uniform magnetic field of $$30\times {10}^{6}T$$ exits in the $$+X$$ direction. A particle of charge $$+e$$ and mass $$1.67\times {10}^{-27}kg$$ is projected through the field in the $$+Y$$ direction with a speed of $$4.8\times {10}^{6}m/s$$

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