Question

A charge $Q$ of mass $M$ moving in a straight line is accelerated by a potential difference $V$ . It enters a uniform magnetic field $B$ perpendicular to its path. Deduce an expression in terms of $V$ for the radius of the circular path in which it travels?

Answer 1

Hint: - Since we know that if there is a charged particle $Q$ that is moving with a velocity $v$ entering in a magnetic field with perpendicular direction, the magnetic field will make the charged particle move in a circular motion. Then we will find the kinetic energy with the help of the velocity $v$ and then equate the kinetic energy with electrical energy to get the answer.
Formula used: ${F_c} = \dfrac{{M{v^2}}}{r}$ , $K.E = \dfrac{1}{2}M{v^2}$

Complete step-by-step answer:
The value of magnetic force will be-
${F_m} = Q\left( {\overrightarrow v \times \overrightarrow B } \right)$
As the uniform magnetic field is perpendicular to the direction of motion of the charge. So,
${F_m} = Q\sin {90^ \circ }vB$
$\Rightarrow {F_m} = QvB$
Since the particle makes a circular path having a radius say $R$, the magnetic force will provide a necessary centripetal force. We know the formula for centripetal force is${F_c} = \dfrac{{M{v^2}}}{R}$. Therefore,
$\dfrac{{M{v^2}}}{R} = QvB$
$\Rightarrow R = \dfrac{{Mv}}{{QB}}$
Rearrange the above equation to get the value of $v$ ,
$v = \dfrac{{QBR}}{M}$ .............. $\left( 1 \right)$
As we know that the formula for kinetic energy is,
$K.E = \dfrac{1}{2}M{v^2}$
Putting the value of $v$ from the equation $\left( 1 \right)$ , we will have;
$K.E = \dfrac{1}{2}M{\left( {\dfrac{{QRB}}{M}} \right)^2}$
$\Rightarrow K.E = \dfrac{1}{2}M\dfrac{{{Q^2}{R^2}{B^2}}}{{{M^2}}}$
On further simplifying we get,
$K.E = \dfrac{{{Q^2}{R^2}{B^2}}}{{2M}}$
An electrical potential $V$ was applied to get the charged particle accelerated. The particle applied some velocity due to which it's kinetic energy is $K.E = \dfrac{{{Q^2}{R^2}{B^2}}}{{2M}}$
Now, this kinetic energy should be equal to electrical energy $QV$ where $V$ is the electric potential.
$\therefore K.E = QV$
$\Rightarrow \dfrac{{{Q^2}{R^2}{B^2}}}{{2M}} = QV$
$\Rightarrow {R^2} = \dfrac{{2MV}}{{Q{B^2}}}$
On taking square root both the sides we get,
$\Rightarrow R = \dfrac{1}{B}\sqrt {\dfrac{{2MV}}{Q}}$
Hence, the radius of the circular path in which the charge is moving in terms of potential difference $V$ is $\dfrac{1}{B}\sqrt {\dfrac{{2MV}}{Q}}$ .

Note: The direction in which the electron starts to move due to the magnetic field can be found using Fleming’s left-hand rule. The index finger refers to the intensity of magnetization, the next finger refers to the direction along which the charged particle is moving. The direction of the thumb indicates the direction along which the charged particle gets bent due to the perpendicular magnetic field.