Q.

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 μN) if the beam current is equal to I=0.80mA. Take mass of proton as 1.67×10-27kg.

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answer is 20.04.

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Detailed Solution

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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