A charged particle of mass m=1g and charge q=1μC enters along AB at point A in a uniform magnetic field B=1.2 T existing in the rectangular region of size a×b, where a=4m and b=3m. The particle leaves the region exactly at corner point C. What is the speed v(in m s-1 )of the particle?

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answer is 5.

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Detailed Solution

Let speed of the particle is v (speed will remain constant)r=AO=CO=mvqB also ar=sinθ,r-br=cosθSolve the above equation to getv=qBa2+b22mb=10-6×1.242+322×10-6×3=5 m/s

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