Question

In an experiment, electrons are accelerated, from rest, by applying a voltage of $500V$ and magnetic field is $100mT$. Find the radius of the path. It is given that charge of electron is $1.6 \times {10^{ - 19}}C$ and mass of the electron is $9.1 \times {10^{ - 31}}kg$
A. $7.5 \times {10^{ - 4}}m$
B. $7.5 \times {10^{ - 3}}m$
C. $7.5m$
D. $7.5 \times {10^{ - 2}}m$

Answer 1

Use the conservation of momentum which states that the momentum remains constant. Also, use the relation between momentum and kinetic energy and consider charge and radius as the charge and radius of the electron respectively.

Complete step by step answer:
Let the initial speed of the electron be $u$ and the final speed be $v$.
According to the question, it is given that –
$u = 0 \\\ v = v \\\ B = 100mT \\\ V = 500V \\\ {m_e} = 9.1 \times {10^{ - 31}}kg \\\ {q_e} = 1.6 \times {10^{ - 19}}C \\\$
Now, we have to use the conservation of momentum which states that momentum on an isolated system remains constant which means it can neither be created nor destroyed. So, if the initial momentum is ${P_i}$ and final momentum is ${P_f}$.
$\therefore \Delta {P_i} = \Delta {P_f} \cdots \left( 1 \right)$
We know that,
$\Delta {P_i} = qrB \\\ \Delta {P_f} = m\left( {v - u} \right) \\\$
Putting these values in equation $\left( 1 \right)$, we get –
$\Rightarrow qrB = mv$ $\left[ {\because u = 0} \right]$
$\Rightarrow r = \dfrac{{mv}}{{qB}}$
We know that momentum is equal to the product of mass and velocity.
$\therefore r = \dfrac{P}{{qB}} \cdots \left( 2 \right)$
Consider the final momentum to be $P$.
Now, use the relation between momentum and kinetic energy which can be expressed as –
$P = \sqrt {2mK}$
As we know that, $K = q\Delta V$
$\therefore P = \sqrt {2mq\Delta V} \cdots \left( 3 \right)$
Putting the value of momentum from equation $\left( 3 \right)$ in equation $\left( 2 \right)$, we get –
$\therefore r = \dfrac{{\sqrt {2mq\Delta V} }}{{qB}}$
Putting the value of charge of electron and mass of electron in the above equation, we get –
$r = \dfrac{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}} \times 500} }}{{1.6 \times {{10}^{ - 19}} \times 100 \times {{10}^{ - 3}}}}$
Now, by further solving we get the value of radius as –
$r = 7.5 \times {10^{ - 4}}m$
Hence, the correct option is (A).

Note: Here most of the times we forget to take the milli, micro factors into consideration and which may lead to different answers. As these kinds of questions will differ in the powers of 10 only.
If a massive particle and a light particle have the same momentum, the light one will have a lot more kinetic energy. If a light particle and a heavy one has the same velocity, the heavy one has more kinetic energy.