Q 12-04-015NEETJEE MainTop questionHard
A proton moves with speed $3 \times 10^5$ m/s at $60^\circ$ to a uniform magnetic field of $0.2$ T. The distance it advances along the field in one revolution (the pitch of its helix) is about ($m_p = 1.67 \times 10^{-27}$ kg, $e = 1.6 \times 10^{-19}$ C)
Answer: (A) $4.9$ cm
Only the component along $B$ carries the proton forward: $v_\parallel = v\cos 60^\circ = 1.5 \times 10^5$ m/s.
Time for one turn: $T = \dfrac{2\pi m}{eB}$.
Pitch $= v_\parallel T = \dfrac{2\pi m v_\parallel}{eB} = \dfrac{2\pi \times 1.67 \times 10^{-27} \times 1.5 \times 10^5}{1.6 \times 10^{-19} \times 0.2} \approx 0.049$ m $= 4.9$ cm.
Solution by Sreeraj P, M.Sc Physics