Question

A beam of cathode rays is subjected to crossed electric (E) and magnetic fields . The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by

• A. $ \frac{{{B}^{2}}}{2V{{E}^{2}}} $
• B. $ \frac{2V{{B}^{2}}}{{{E}^{2}}} $
• C. $ \frac{2V{{E}^{2}}}{{{B}^{2}}} $
• D. $ \frac{{{E}^{2}}}{2V{{B}^{2}}} $

Answer 1

Answer: (D)

As the electron beam is not deflected, then $ {{F}_{m}}={{F}_{e}} $ or $ Bev=Ee $ or $ v=\frac{E}{B} $ ... (i) As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If V is the potential difference between the anode and cathode, then potential energy of the electron at cathode = eV. Also, kinetic energy of the electron at anode $ =\frac{1}{2}m{{v}^{2}}. $ According to law of conservation of energy $ \frac{1}{2}m{{v}^{2}}=eV $ or $ v=\sqrt{\frac{2eV}{m}} $ ?(ii) From Eqs. (i) and (ii), we have $ \sqrt{\frac{2eV}{m}}=\frac{E}{B} $ or $ \frac{e}{m}=\frac{{{E}^{2}}}{2V{{B}^{2}}} $
Note : - (where V is the potential difference between cathode and anode)