Proton with kinetic energy of $$1$$ MeV moves from south to north. It gets an acceleration of $$1012m/s2$$ by an applied magnetic field (west$$ to $$east). The value of magnetic field: (Rest$$ mass of proton is $$1.6$$×$$10$$−$$27kg)

Question

Proton with kinetic energy of $$1$$ MeV moves from south to north. It gets an acceleration of  $$1012m/s2$$ by an applied magnetic field (west$$ to $$east). The value of magnetic field: (Rest$$ mass of proton is  $$1.6$$×$$10$$−$$27kg)

Infinity Learn · Accepted Answer

∵K.E of $$proton=12mV2$$ ⇒$$1$$×$$106$$×$$1.6$$×$$10$$−$$19=12$$×$$1.6$$×$$10$$−$$27$$×$$V2$$⇒$$V=2$$×$$107$$ $$m/s$$∴   Magnetic Force accelerates the proton ∴$$BqV=$$ ma  ⇒$$B=maqV$$  ⇒$$B=1.6$$×$$10$$−$$27$$×$$10121.6$$×$$10$$−$$19$$×$$2$$×$$107$$                   $$=0.71$$×$$10$$−$$3TSo$$,  B $$=$$ $$0.71mT$$

Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of $10^{12} m / s^{2}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{- 27} k g)$

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Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 1012m/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6×10−27kg)

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Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of $10^{12} m / s^{2}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{- 27} k g)$