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 bysin a = Bd(q/2Vm)where m & q are the mass & charge of the particle respectively
Radhika Batra , 10 Years ago
Grade 11
1 Answers
ROSHAN MUJEEB
Last Activity: 4 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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