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A proton accelerated by a potential dif...

A proton accelerated by a potential difference `V = 500 kV` flies through a uniform transverse magnetic filed the induction `B = 0.54 T`. The field occupies a region of space `d = 10 cm` in thickness (Fig). Find the angle `alpha` through which the proton deviates from the initial direction of its motion.

A

`15^(@)`

B

`30^(@)`

C

`45^(@)`

D

`60^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B

According to following figure `sintheta=(d)/(r )`

Also, `r=sqrt(2mK)/(qB)=(1)/(B)sqrt((2mV)/(q))`
`:. Sintheta=Bdsqrt((q)/(2mV))`
`=0.51xx0.1sqrt((1.6xx10^-19)/(2xx1.67xx10^-27xx500xx10^3))`
` (1)/(2)impliestheta=30^(@)`
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A2Z-SOURCE AND EFFECT OF MAGNETIC FIELD-Section D - Chapter End Test
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  10. A proton moving with a constant velocity passes through a region of sp...

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  11. A charged particle fo mass m and charge q enters along AB at point A ...

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  15. A proton , a deutron and alpha )- particle having the same kinetic en...

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  16. Three long straight and parallel wires carrying currents are arranged...

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  17. Two straight horizontal parallel wires are carrying the same current i...

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  18. A proton and an alpha-particle enter a uniform magnetic field moving w...

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  19. A proton (mass = 1.67xx10^(-27) kg and charge 1.6xx10^(-19) C) enters ...

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  20. An alpha particle travels in a circular path of radius 0.45 m in a mag...

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  21. A unifrom magnetic field B is acting from south to north and is of m...

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