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  • A beam of protons with velocity <img alt="4 \times 10^{5} \mathrm{~m} / \mathrm{s}" src="https://entrancecorner.oncodecogs.com/gif.latex?4%20%5Ctimes%2010%5E%7B5%7D%20%5Cmathrm%7B%7Em%7D%20/%2

A beam of protons with velocity 4 \times 10^{5} \mathrm{~m} / \mathrm{s} enters a uniform magnetic field of 0.4 T at an angle of 60 to the magnetic field. Find the radius of the helical path of the proton beam and the time period of revolution-

Option: 1

1.63 \times 10^{-7}


Option: 2

1.93 \times 10^{-6} \mathrm{~s}


Option: 3

2.63 \times 10^{-7} s


Option: 4

none


Answers (1)

best_answer

 The component of velocity perpendicular to the field is v \sin \theta, the radius of the helical path,

\begin{aligned} r=\frac{m v \sin \theta}{2 B} & =\frac{\left(1.67 \times 10^{-27}\right)\left(4 \times 10^{5}\right) \sqrt{3}}{\left(1.6 \times 10^{-19}\right) \times 0.4 \times 2} \\ & =9 \times 10^{-3} \mathrm{~m}=0.9 \mathrm{~cm} . \end{aligned}

Time period of revolution,

T=\frac{2 \pi r}{v \sin \theta}=\frac{2 \times 3.14 \times 0.009 \times 2}{4 \times 10^{5} \times \sqrt{3}}=1.63 \times 10^{-7} \mathrm{~s}

 

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shivangi.bhatnagar

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