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Question

A proton accelerated by a potential $$V=500$$ $$KV$$ moves through a transverse magnetic field $$B=0.51$$ $$T$$ as shown in the figure. Then, the angle $$\theta $$ through which the proton deviates from the initial direction of its motion is (approximately) 

23899_2584806518cc4cf79df0da45ce684803.png


A
15o
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B
30o
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C
45o
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D
60o
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Solution

The correct option is A $$30^{o}$$
Proton is accelerated by a potential difference $$V=500kV$$
So,
$$\dfrac{1}{2}{mv}^2=qV$$
$$v=\sqrt{\dfrac{2qV}{m}}$$
From the figure
$$d=r \sin\theta$$
$$10\times10^{-2}=\sin\theta\times\dfrac{mv}{qB}$$
$$10^{-1} =\dfrac{m\sin\theta}{qB} \sqrt{\dfrac{2qV}{m}}$$
$$\sin\theta=0.51$$
$$\theta=30^{\circ}$$

Physics

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