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 (2Kcos$$θ$$)/(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?

Question

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 (2Kcos$$θ$$)/(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?

Infinity Learn · Accepted Answer

$$r=mvqB=2KqvB$$  Thus, $$OA=rcos$$θ (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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