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(C) When electrons pass through crossed electric $(E)$ and magnetic $(B)$ fields undeflected,the electric force equals the magnetic force: $eE = evB$,

Vedclass · Accepted Answer

$1.80 	imes 10^{11} \ C \ kg^{-1}$

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(C) When electrons pass through crossed electric $(E)$ and magnetic $(B)$ fields undeflected,the electric force equals the magnetic force: $eE = evB$,which implies $v = E/B$.The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $\frac{1}{2}mv^2 = eV$,so $v^2 = \frac{2eV}{m}$.Substituting $v = E/B$ into the equation,we get $(E/B)^2 = \frac{2eV}{m}$,which simplifies to $\frac{e}{m} = \frac{E^2}{2VB^2}$.Given values: $E = 3.6 	imes 10^4 \ Vm^{-1}$,$B = 1.2 	imes 10^{-3} \ T$,$V = 2.5 	imes 10^3 \ V$.Calculating $\frac{e}{m} = \frac{(3.6 	imes 10^4)^2}{2 	imes (2.5 	imes 10^3) 	imes (1.2 	imes 10^{-3})^2} = \frac{12.96 	imes 10^8}{5000 	imes 1.44 	imes 10^{-6}} = \frac{12.96 	imes 10^8}{7.2 	imes 10^{-3}} = 1.8 	imes 10^{11} \ C \ kg^{-1}$.

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