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A proton with kinetic energy 1·3384 × 10⁻¹⁴ J moving horizontally from north to south, enters a uniform magnetic field B of 2·0 mT directed eastward. Calculate : (a) the speed of the proton (b) the magnitude of acceleration of the proton (c) the radius of the path traced by the proton [Take (q/m) for proton = 1·0 × 10⁸ C/kg]
Assertion (A) : An electron and a proton enter with the same momentum p in a magnetic field B such that p ⊥ B. Then both describe a circular path of the same radius. Reason (R) : The radius of the circular path described by the charged particle (charge q, mass m) moving in the magnetic field B is given by r = mv/qB.
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