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Q.

A proton beam passes without deviation through a region of space where there are uniform transverse mutually perpendicular electric and magnetic fields with E=120 kV/m and B=50 mT. Then the beam strikes a grounded target. Find the force ( in ×10-5 N) imparted by the beam on the target if the beam current is equal to I = 0.80 mA.

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Detailed Solution

Fe=Fm or eE=eBv ∴ v=EB=120×10350×10-3=2.4×106 m/s Let n be the number of protons striking per second.  Then, ne=0.8×10-3 or  n=0.8×10-31.6×10-19=5×1015 m/s Force imparted: Rate of change of momentum = nmv =5×1015×1.67×10-27×2.4×106 =2.0×10-5 N
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A proton beam passes without deviation through a region of space where there are uniform transverse mutually perpendicular electric and magnetic fields with E=120 kV/m and B=50 mT. Then the beam strikes a grounded target. Find the force ( in ×10-5 N) imparted by the beam on the target if the beam current is equal to I = 0.80 mA.