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A magnetic field B exists between OA and...

A magnetic field B exists between OA and OB. Inclined at an angle `(theta)`, a charged particle strikes at point A on surface OA, at a distance `(2K cos(theta))(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?

A

`theta`

B

`90^(@)-theta`

C

`90^(@)+theta`

D

`90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
`r=(mv)/(qB)=(2K)/(qvB)`
Thus, `OA = r cos theta (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^(@)`. Particle leaves circular path at C. thus, required angle `90^(@)`.
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