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