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)

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

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More