Question

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

Answer 1

Answer:

$\sqrt{3}R$

Explanation:

Case 1:

$\Delta V$ = potential difference through which the charge is accelerated = V

$m$ = mass of the charge

$q$ = magnitude of charge

$v$ = speed gained by the charge due to potential difference

Using conservation of energy

Kinetic energy gained = Electric potential energy lost

$(0.5)mv^{2}=q\Delta V$

$v = \sqrt{\frac{2q\Delta V}{m}}$

$B$ = magnitude of magnetic field

$R$ = Radius of arc

Radius of arc is given as

$R = \frac{mv}{qB}$

$R = \frac{m}{qB}\sqrt{\frac{2q\Delta V}{m}}$

$R = \frac{1}{B}\sqrt{\frac{2m V}{q}}$                                eq-1

Case 2:

$\Delta V'$ = potential difference through which the charge is accelerated = 3V

$m$ = mass of the charge

$q$ = magnitude of charge

$v'$ = speed gained by the charge due to potential difference

Using conservation of energy

Kinetic energy gained = Electric potential energy lost

$(0.5)mv'^{2}=q\Delta V'$

$v' = \sqrt{\frac{2q\Delta V'}{m}}$

$B$ = magnitude of magnetic field

$R'$ = Radius of arc

Radius of arc is given as

$R' = \frac{mv'}{qB}$

$R' = \frac{m}{qB}\sqrt{\frac{2q\Delta V'}{m}}$

$R' = \frac{1}{B}\sqrt{\frac{2m\Delta V'}{q}}$

$R' = \frac{1}{B}\sqrt{\frac{2m (3V)}{q}}$

$R' = \sqrt{3}\frac{1}{B}\sqrt{\frac{2m V}{q}}$

Using eq-1

$R' = \sqrt{3}R$

Answer 2

The radius of the circular arc is $\sqrt{3}$R.The radius is half of the diameter of the circle and the diameter is the smallest chord. It is a chord that departs via the center of the circle.

What is kinetic energy?

The kinetic energy (KE) is defined as the one-half of the mass of each gas molecule times multiplied by the square of velocity.

According to the law of conservation of energy;

The kinetic energy gained = Electric potential energy

$\rm (0.5) mv^2= q \triangle V \\\\ \rm v= \sqrt{\frac{2q \triangle V}{y} }$

The radius arc is found by;

$\rm R= \frac{mv}{qB}$

$\rm R = \frac{m}{qB} \sqrt{\frac{2q \triangle v}{m} }$

$\rm R= \frac{1}{B} \sqrt{\frac{2mv}{q} }$

Where,

Potential difference through which the charge is accelerated = V

Mass of the charge = m

The magnitude of the charge=q

The speed gained by the charge due to potential difference=v

The magnitude of magnetic field=B

The radius of arc = R

Condition-2

The gain in kinetic energy = lost in the electric potential

$\rm (0.5) mv'^2= q \triangle v' \\\\ \rm v'= \sqrt{\frac{2q \triangle v'}{y} }$

The radius of the arc is;

$\rm R'= \frac{mv'}{qB} \\\\ \rm R' = \frac{m}{qB} \sqrt{\frac{2q \triangle v'}{m} } \\\\ \rm R= \frac{1}{B} \sqrt{\frac{2mv'}{q} }$

$\rm R'= \frac{1}{B} \sqrt{\frac{2m(3v)}{q} }\\\\ \rm R'=\sqrt{3} \frac{1}{B}\sqrt{ \frac{2mv}{q} } \\\\ \rm R'=\sqrt{3} R$

Hence the radius of the circular arc is $\sqrt{3}$R

To learn more about kinetic energy refer to the link ;

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