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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 $(2 K \cos θ) / (q v B)$ 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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$θ$

$90 ° + θ$

$90 °$

$90 ° - θ$

answer is D.

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Detailed Solution

$r = \frac{m v}{q B} = \frac{2 K}{q v B} Thus, O A = r \cos θ (given)$

On line OB, centre O' of part of circle (in which particle moves) lies. The tangent at C with OB makes an angle of 90^o. Particle leaves circular path at C. Thus, required angle is 90^o.

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