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  • A proton moving with a velocity  <img alt="3 \times 10^{\circ} \mathrm{m} / \mathrm{s}" src="https://entrancecorner.oncodecogs.com/gif.latex?3%20%5Ctimes%2010%5E%7B%5Ccirc%7D%20%5Cmathrm%

A proton moving with a velocity  3 \times 10^{\circ} \mathrm{m} / \mathrm{s}  enters a magnetic field of 0.3 tesla at an angle of 30^{\circ}  with the field. The radius of curvature of its path will be \mathrm{(\mathrm{e} / \mathrm{m} ~for~ proton~=10^8 \mathrm{C} / \mathrm{kg} )}

Option: 1

2 cm


Option: 2

0.5 cm


Option: 3

0.02 cm


Option: 4

1.25 cm


Answers (1)

best_answer

\mathrm{r}=\frac{\mathrm{mv} \sin \theta}{\mathrm{Be}}=\frac{3 \times 10^5 \sin 30^{\circ}}{0.3 \times 10^8} \\

\frac{3 \times 10^5 \times \frac{1}{2}}{3 \times 10^7}=0.5 \times 10^{-2} \mathrm{~m}=0.5 \mathrm{~cm} .
 

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