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Q.

A proton accelerated by a potential difference 500 kV moves though a transverse magnetic field of 0.51 T as shown in figure. The angle θ through which the proton deviates from the initial direction of its motion, is

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a

150

b

300

c

450

d

600

answer is B.

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Detailed Solution

According to following figure, sin θ= drAlso,  r= 2mKqB  =  1B 2mVq∴     sin θ=Bd q2mV=0.51  ×  0.1  1.6  ×  10−192  × 1.67   × 10−27 × 500   × 103=12  ⇒  θ= 300
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A proton accelerated by a potential difference 500 kV moves though a transverse magnetic field of 0.51 T as shown in figure. The angle θ through which the proton deviates from the initial direction of its motion, is