A proton beam passes without deviation through a region of space where there are uniform transverse mutually perpendicular electric and magnetic field with $$E=120kVm$$ and $$B=50$$ mT. Then, the beam strikes a grounded target. Find the force imparted by the beam on the target (in$$ μ$$N) if the beam current is equal to $$I=0.80mA$$. Take mass of proton as $$1.67$$×$$10-27kg$$.

Question

A proton beam passes without deviation through a region of space where there are uniform transverse mutually perpendicular electric and magnetic field with $$E=120kVm$$ and $$B=50$$ mT. Then, the beam strikes a grounded target. Find the force imparted by the beam on the target (in$$ μ$$N) if the beam current is equal to $$I=0.80mA$$. Take mass of proton as $$1.67$$×$$10-27kg$$.

Infinity Learn · Accepted Answer

$$Fe=Fm$$ or $$eE=eBv$$∴ $$v=EB=120$$×$$10350$$×$$10$$−$$3=2.4$$×$$106m/sLet$$ n be the number of protons striking per second. Then, $$ne=0.8$$×$$10$$−$$3or$$ $$n=0.8$$×$$10$$−$$31.6$$×$$10$$−$$19=5$$×$$1015m/sForce$$ imparted (F) $$=$$ Rate of change of momentum $$=$$ $$nmvF=5$$×$$1015$$×$$1.67$$×$$10$$−$$27$$×$$2.4$$×$$106=20.04$$×$$10$$−$$6N=20.04$$μN

Motion of a charged particle

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Motion of a charged particle

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Fe=Fm or eE=eBv∴ v=EB=120×10350×10−3=2.4×106m/sLet n be the number of protons striking per second. Then, ne=0.8×10−3or n=0.8×10−31.6×10−19=5×1015m/sForce imparted (F) = Rate of change of momentum = nmvF=5×1015×1.67×10−27×2.4×106=20.04×10−6N=20.04μN

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