Question: 3. An electron (mass 9 \times 10^{-31}\ \mathrm{kg} and charge 1.6 \times 10^{-19}\ \mathrm{C}) moving with speed c/100 (c= speed of light) is injected into a magnetic field \vec{B} of magnitude 9 \times 10^{-4}\ \mathrm{T} perpendicular to its direction of motion. We wish to apply an uniform electric \vec{E} together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c=3 \times 10^{8}\ \mathrm{m\ s^{-1}})
(1) \vec{E} is perpendicular to \vec{B} and its magnitude is 27 \times 10^{4}\ \mathrm{V\ m^{-1}}
(2) \vec{E} is perpendicular to \vec{B} and its magnitude is 27 \times 10^{2}\ \mathrm{V\ m^{-1}}
(3) \vec{E} is parallel to \vec{B} and its magnitude is 27 \times 10^{2}\ \mathrm{V\ m^{-1}}
(4) \vec{E} is parallel to \vec{B} and its magnitude is 27 \times 10^{4}\ \mathrm{V\ m^{-1}}
Answer: Option (2)
Explanation:
For no deflection the net transverse force on the electron must be zero, so electric force balances magnetic force:
q\vec{E} + q(\vec{v}\times\vec{B})=\vec{0}Magnitude condition (and using signs later) gives:
E = vBFirst find the speed:
v=\dfrac{c}{100}=\dfrac{3\times10^{8}}{100}=3\times10^{6}\ \mathrm{m\ s^{-1}}Now compute E:
E = vB = \bigl(3\times10^{6}\bigr)\bigl(9\times10^{-4}\bigr)=27\times10^{2}\ \mathrm{V\ m^{-1}}Direction: \vec{E} must be perpendicular to \vec{B} (and chosen so q\vec{E} cancels q\vec{v}\times\vec{B}); since the charge is negative the required \vec{E} is opposite to \vec{v}\times\vec{B}.
Hence option (2) is correct.