In Thomson's experiment to determine $\frac{e}{m}$ of an electron, it is found that an electron beam having…

In Thomson's experiment to determine $\frac{e}{m}$ of an electron, it is found that an electron beam having a kinetic energy of $45.5 \mathrm{eV}$ remains undeflected, when subjected to crossed electric and magnetic fields. If $E=1 \times 10^3 \mathrm{Vm}^{-1}$, the value of $B$ is (mass of the electron is $9.1 \times 10^{-31} \mathrm{~kg}$ )
  1. $2.5 \times 10^{-3} \mathrm{~Wb} \mathrm{~m}^{-2}$
  2. $5.0 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$
  3. $2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$
  4. $1.0 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$

Solution

The kinetic energy $K=\frac{1}{2} m v^2$ $v^2=\frac{2 K}{m}$ $v^2=\frac{2 \times 45.5 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $v^2=16 \times 10^{12}$ $v=4 \times 10^6$ Again velocity, $v=\frac{E}{B}$ $4 \times 10^6=\frac{1 \times 10^3}{B}$ $B=\frac{1 \times 10^3}{4 \times 10^6}$ $B=2.5 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-2}$

Asked in: AP EAMCET 2010

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