Question

A charge particle after being accelerated through a potential difference V
enters in a uniform magnetic field and moves in a circle of radius r. If V is
doubled, the radius of the circle will become
(a) 2r
(b) root2r
(c) 4r
(d) r/root2​

Answer 1

Answer:

the charged particle after being accelerated through a potential difference v the radius of circle will become root 2r

Answer 2

The new radius of the circle is $\sqrt{2}r$.

(b) is correct option.

Explanation:

Given that,

Potential difference = V

New potential difference = 2 V

The speed when accelerated through potential difference V is

By conservation of energy,

The kinetic energy equal to the change in potential difference.

$qV=\dfrac{1}{2}mv^2$

$v^2=\dfrac{2qV}{m}$...(I)

We need to calculate the new radius of the circle

Using relation magnetic force and centripetal force

$F_{B}=F_{c}$

$qvB=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$.....(II)

Put the value of v in equation (II)

$r=\dfrac{m\times\sqrt{\dfrac{2qV}{m}}}{qB}$

$r=\dfrac{\sqrt{2Vm}}{B\sqrt{q}}$

If the potential difference is V.

The radius is,

$r=\dfrac{\sqrt{2Vm}}{B\sqrt{q}}$....(III)

If the new potential difference is 2V.

The new radius is,

$r'=\dfrac{\sqrt{2(2V)m}}{B\sqrt{q}}$.....(IV)

Divided equation (IV) by equation (III)

$\dfrac{r'}{r}=\dfrac{\dfrac{\sqrt{2(2V)m}}{B\sqrt{q}}}{\dfrac{\sqrt{2Vm}}{B\sqrt{q}}}$

$\dfrac{r'}{r}=\sqrt{\dfrac{2V}{V}}$

$r'=\sqrt{2}r$

Hence, The new radius of the circle is $\sqrt{2}r$.

Learn more :

Topic : potential difference

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