A proton accelerated by a potential difference 500 kV flies through a uniform transverse magnetic field 0 . 1 T . The field is spread on a region of 1 . 0 cm thickness. The angle through which the proton gets deviated from its original direction is (Proton mass = 1 . 6 × 10 - 27 kg and charge of proton = 1 . 6 × 10 - 19 C )

Question

A proton accelerated by a potential difference 500 kV flies through a uniform transverse magnetic field 0 . 1 T . The field is spread on a region of 1 . 0 cm thickness. The angle through which the proton gets deviated from its original direction is (Proton mass = 1 . 6 × 10 - 27 kg and charge of proton = 1 . 6 × 10 - 19 C ), 0 . 01 rad, 0 . 1 rad, 0 . 05 rad, 0 . 08 rad

MARKS App · Accepted Answer

The radius of circular path is, R = 2 m K E q B = 2 × 1 . 6 × 10 - 27 × 1 . 6 × 10 - 16 × 500 × 10 3 1 . 6 × 10 - 16 × 0 . 1 = 1 m Since K E = q V For small value of θ and arc of cirlcle, θ = a r c R = 1 × 10 - 2 1 = 10 - 2 = 0 . 01 rad

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