A magnetic field B exists between OA and OB. Inclined at an angle q, a charged particle strikes at point A on surface OA,at a distance (2Kcosθ)/(qvB) from O, where K is kinetic energy, q is the charge and v is the velocity of the particle.At what angle with horizontal (measured from end B) will the charged particle emerge from OB?
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a
θ
b
90°-θ
c
90°+θ
d
90°
answer is D.
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Detailed Solution
r=mvqB=2KqvB Thus, OA=rcosθ (given) On line OB, centre O' of part of circle (in which particle moves) lies. The tangent at C with OB makes an angle of 90o. Particle leaves circular path at C. Thus, required angle is 90o.
A magnetic field B exists between OA and OB. Inclined at an angle q, a charged particle strikes at point A on surface OA,at a distance (2Kcosθ)/(qvB) from O, where K is kinetic energy, q is the charge and v is the velocity of the particle.At what angle with horizontal (measured from end B) will the charged particle emerge from OB?