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(C) In the Thomson experiment,the condition for no deflection of a charged particle moving through crossed electric field $E$ and magnetic field $B$ i

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$B$ is increased $14.4$ times

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(C) In the Thomson experiment,the condition for no deflection of a charged particle moving through crossed electric field $E$ and magnetic field $B$ is given by the balance of forces: $qE = qvB$,which implies $v = E/B$.The kinetic energy gained by the particle accelerated through a potential difference $V$ is $qV = \frac{1}{2}mv^2$,which gives $v = \sqrt{\frac{2qV}{m}}$.Equating the two expressions for velocity: $\frac{E}{B} = \sqrt{\frac{2qV}{m}}$.Squaring both sides: $\frac{E^2}{B^2} = \frac{2qV}{m}$,which leads to $\frac{q}{m} = \frac{E^2}{2VB^2}$.For a fixed potential $V$ and electric field $E$,we have $B^2 \propto \frac{1}{m}$.If the mass $m$ is increased by a factor of $208$,then $B^2$ must change by a factor of $1/208$.Therefore,$B$ must change by a factor of $\sqrt{1/208}$,meaning $B$ must be multiplied by $\sqrt{208} \approx 14.4$ to maintain the condition.

In the Thomson experiment for finding $e/m$ for electrons,the beam of electrons is replaced by a beam of muons (particles with the same charge as electrons but with a mass $208$ times that of electrons). The condition for no deflection is satisfied if:

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