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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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