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(C) The magnetic force on a moving charge is given by $F = qvB \sin 	heta$. For a transverse magnetic field,$	heta = 90^{\circ}$,so $F = qvB$. The v

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$\sqrt{2}F$

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(C) The magnetic force on a moving charge is given by $F = qvB \sin 	heta$. For a transverse magnetic field,$	heta = 90^{\circ}$,so $F = qvB$. The velocity $v$ of an electron accelerated through a potential difference $V$ is given by $\frac{1}{2}m_e v^2 = eV$,which implies $v = \sqrt{\frac{2eV}{m_e}}$. Substituting this into the force equation: $F = e \left( \sqrt{\frac{2eV}{m_e}} ight) B$. This shows that $F \propto \sqrt{V}$. When the potential is increased to $2V$,the new force $F'$ is: $F' = e \left( \sqrt{\frac{2e(2V)}{m_e}} ight) B = \sqrt{2} \cdot e \left( \sqrt{\frac{2eV}{m_e}} ight) B = \sqrt{2}F$.

An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

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