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Q.

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.
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