A beam of protons with speed 4 × 105 ms–1 enters a uniform magnetic field of 0.3T at an angle of 60º to the magnetic field. the pitch of the resulting helical path of protons is close to : (Mass of the proton = 1.67 × 10–27 kg, charge of the proton = 1.69 × 10–19 C)
(1) 12 cm
(2) 2 cm
(3) 4 cm
(4) 5 cm
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Correct answer is option 3) 4cm
Explaination::
\(Pitch=(V\cos\theta)T\)
\(=(V\cos\theta)\frac{2\pi m}{eB}\)
\(=(4\times10^5\cos60°){2\pi\over0.3\times10}\left({1.67\times10^{-27}\over1.69\times10^{19}}\right)\)
= 4cm
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