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The figure shows a region of length ‘l’ with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity 4 ×105 ms–1 making an angle 60° with the field. If the proton completes 10 revolution by the time it cross the region shown, ‘l’ is close to (mass of proton = 1.67 × 10–27 kg, charge of the proton = 1.6 × 10–19 C)

1) 0.11 m 2) 0.22 m 3) 0.44 m 4) 0.88 m Answer: (3) l = 10 × pitch = 10 × vcos 60° ×(2π m/qB) = 10 × v ×(1/2) ×(2πm/qB)... View Article