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(C) The magnetic force on a charged particle moving in a magnetic field is given by $F = Bqv \sin 	heta$. Since the electron enters the field transve

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$\sqrt{2} F$

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(C) The magnetic force on a charged particle moving in a magnetic field is given by $F = Bqv \sin 	heta$. Since the electron enters the field transversely,$	heta = 90^\circ$,so $F = Bqv$.The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $\frac{1}{2}mv^2 = eV$,which implies $v = \sqrt{\frac{2eV}{m}}$.Substituting this into the force equation: $F = B e \sqrt{\frac{2eV}{m}} = B e \sqrt{\frac{2e}{m}} \sqrt{V}$.From this expression,we see that $F \propto \sqrt{V}$.If the potential is increased to $V' = 2V$,the new force $F'$ will be $F' \propto \sqrt{2V}$.Taking the ratio: $\frac{F'}{F} = \frac{\sqrt{2V}}{\sqrt{V}} = \sqrt{2}$.Therefore,$F' = \sqrt{2} F$.

An electron accelerated through a potential difference $V$ enters 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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