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An electron emitted by heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity , (b) makes an angle of 30^@ with the initial velocity. |
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Answer» Solution :When accelerated through a potential difference of `V = 2.0 kV = 2000 V,` the electron will aquire a speed given by `v = sqrt((2eV)/(m_e)) = sqrt((2 xx 1.6 xx 10^(-19) xx 2000)/(9.1 xx 10^(-31))) = 2.65 xx 10^(7) ms^(-1)` (a) When the MAGNETIC field B = 0.15 T is TRANSVERSE to the initial velocity of electron, its trajectory will be a circle of radius `r = (m_e v)/(eB) = (9.1 xx 10^(-31) xx 2.65 xx 10^(7))/(1.6 xx 10^(-19) xx 0.15) = 10^(-3) m or 1 mm`. The circular path is in a plane perpendicular to direction of `vecB`. (b) When the magnetic field `vecB` makes an angle `theta = 30^@` with the initial velocity, the electron will describe a HELIX path of radius `r. = (m_e v sin theta)/(e B) = r sin theta = 10^(-3) xx sin 30^@ = 0.5 xx 10^(-3) m " or " 0.5 mm` and a FORWARD velocity of `v cos 30^@ = 2.65 xx 10^(7) xx cos 30^@ = 2.3 xx 10^(7) ms^(-1)` along `vecB`. |
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