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A particle is accl. thru a uniform transverse magnetic field B which occupies a region of space d in thickness. Prove that the angle 'a' thru which the particle deviates from the initial direction of its motion is given by sin a = Bd (q/2Vm) where m & q are the mass & charge of the particle respectively


A particle is accl. thru a uniform transverse magnetic field B which occupies a region of space d in thickness. Prove that the angle 'a' thru which the particle deviates from the initial direction of its motion is given by                        
                 sin a = Bd(q/2Vm)
where m & q are the mass & charge of the particle respectively


Grade:11

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
3 years ago
According to following figuresinθ=d/r
Also,r=2mK−−−−√qB=1B2mVq−−−−−√r=2mKqB=1B2mVq
∴sinθ=Bdq2mV−−−−−√∴sinθ=Bdq2mV
=0.51×0.11.6×10−192×1.67×10−27×500×103−−−−−−−−−−−−−−−−−−−−−−−−√=0.51×0.11.6×10-192×1.67×10-27×500×103
12⇒θ=30∘

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