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PhysicsNEET 2025Force on a moving charge and on a current

Moving Charges and Magnetism

An electron (mass 9×10−319 \times 10^{-31} kg and charge 1.6×10−191.6 \times 10^{-19} C) moving with speed c/100c/100 (cc = speed of light) is injected into a magnetic field B⃗\vec{B} of magnitude 9×10−49 \times 10^{-4} T perpendicular to its direction of motion. We wish to apply a uniform electric field E⃗\vec{E} together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c=3×108 m s−1c = 3 \times 10^{8}\ \mathrm{m\,s^{-1}}):

  1. A.E⃗\vec{E} is perpendicular to B⃗\vec{B} and its magnitude is 27×104 V m−127 \times 10^{4}\ \mathrm{V\,m^{-1}}
  2. B.E⃗\vec{E} is perpendicular to B⃗\vec{B} and its magnitude is 27×102 V m−127 \times 10^{2}\ \mathrm{V\,m^{-1}}✓
  3. C.E⃗\vec{E} is parallel to B⃗\vec{B} and its magnitude is 27×102 V m−127 \times 10^{2}\ \mathrm{V\,m^{-1}}
  4. D.E⃗\vec{E} is parallel to B⃗\vec{B} and its magnitude is 27×104 V m−127 \times 10^{4}\ \mathrm{V\,m^{-1}}

Solution

For no deflection the electric force must cancel the magnetic force: eE=evBeE = evB, with E⃗\vec{E} perpendicular to both v⃗\vec{v} and B⃗\vec{B}. E=vB=3×106×9×10−4=2.7×103=27×102E = vB = 3\times10^{6} \times 9\times10^{-4} = 2.7\times10^{3} = 27\times10^{2} V/m.

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