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A charge is accelerated from rest through a potential difference V and then enters a uniform magnetic field oriented perpendicular to its path. The magnetic field deflects the particle into a circular arc of radius R. If the accelerating potential is tripled to 3V, the radius of the circular arc will now be:

Respuesta :

Answer:

New radius of the charge particle when potential is increased by 3times of initial value

[tex]R' = \sqrt3 R[/tex]

Explanation:

As we know that charge particle is accelerated due to potential difference V then we have

[tex]\frac{1}{2}mv^2 = qV[/tex]

now the speed of the charge particle is given as

[tex]v = \sqrt{\frac{2qV}{m}}[/tex]

now in constant magnetic field which is perpendicular to the motion of charge  we have

[tex]qvB = \frac{mv^2}{R}[/tex]

now we have

[tex]R = \frac{mv}{qB}[/tex]

now we have

[tex]R = \frac{m}{qB} \times \sqrt{\frac{2qV}{m}}[/tex][/tex]

[tex]R = \frac{1}{B} \sqrt{\frac{2mV}{q}}[/tex]

now if we changed the potential to three times of initial value then we have

[tex]R' = \frac{1}{B} \sqrt{\frac{2m(3V)}{q}}[/tex]

so we have

[tex]\frac{R}{R'} = \frac{1}{\sqrt3}[/tex]

[tex]R' = \sqrt3 R[/tex]

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