Get Answers to all your Questions

header-bg qa

A Proton of mass 1.67 \times 10^{-27} \mathrm{~kg} and charge 1.6 \times 10^{-19} \mathrm{c} is projected with a speed of 2 \times 10^6 \mathrm{~m} / \mathrm{s} at an angle of 60^{\circ} to the x-axis. if a uniform magnetic field of 0.104 T is applied along the y-axis, the path of the Proton is 

Option: 1

A circle of radius 0.2 \mathrm{~m} and time period \pi \times 10^{-7} \mathrm{sec}.


Option: 2

A circle of radius 0.1 \mathrm{~m} and time period 2 \pi \times 10^{-7} \mathrm{sec}.


Option: 3

A helix of radius 0.1 \mathrm{~m} and time period 2 \pi \times 10^{-7} \mathrm{sec}.


Option: 4

A helix of radius 0.2 \mathrm{~m} and time Period 4 \pi \times 10^{-7} \mathrm{sec}.


Answers (1)

best_answer

\because \vec{V} is not Parallel to \vec{B}

\therefore Path of the Proton is helical.

\text { radius } =\frac{m v_{\perp}}{q B} \\

              =0.1 \mathrm{~m} \\

T =\frac{2 \pi m}{q B}=2 \pi \times 10^{-7} \mathrm{sec}

Posted by

HARSH KANKARIA

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE