Question

A proton goes undeflected in a crossed electric and magnetic field (the field are perpendicular to each other) at speed of ${10^5}\,{\text{m}}\,{{\text{s}}^{ - 1}}$. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius $2\,{\text{cm}}$. Find the magnitude of the electric field and the magnetic fields. Take the mass of the proton equals $1.6 \times {10^{ - 27}}\,{\text{kg}}$.

Answer 1

First of all, we will equate the electric field and the magnetic field and find a ration between these two fields. After that we will use the expression of the radius of the path of a particle in a magnetic field to calculate the strength of the field. Lastly, we will find the electric field too, using the relation.

Formula used:
The formula which gives the radius of the path of the subatomic particle is:
$r = \dfrac{{mv}}{{qB}}$ …… (1)
Where,
$r$ indicates the radius of the path.
$m$ indicates the mass of the proton.
$v$ indicates the velocity of the proton.
$B$ indicates the magnetic field.
$q$ indicates the charge

Complete step by step answer:
In the given question, we are supplied with the following information.
There is a proton which goes undeflected in a cross electric and magnetic field. It is given that the two fields are perpendicular to each other. The velocity of the proton is ${10^5}\,{\text{m}}\,{{\text{s}}^{ - 1}}$. The velocity vector is perpendicular to both the fields. The proton moves in a circle of radius of $2\,{\text{cm}}$ when the electric field is switched off. We are asked to find the magnitude of the electric field and the magnetic fields.

To begin with, we will have to first work on comparing the two forces i.e. force of electric field and magnetic field. Since, we are given that the proton goes undeflected under the combined effect of electric and magnetic fields. So, the forces applied by the electric and magnetic fields are equal.
We can write:
q \times E = q \times v \times B \\\ \Rightarrow E = v \times B \\\
As soon as the electric field is switched off, the proton moves by the sole effect of the magnetic field. So, we can write from the equation (1),
r = \dfrac{{mv}}{{qB}} \\\ \Rightarrow B = \dfrac{{mv}}{{qr}} \\\
Now, we substitute the required values in the above expression:
B = \dfrac{{mv}}{{qr}} \\\ \Rightarrow B = \dfrac{{1.6 \times {{10}^{ - 27}} \times {{10}^5}}}{{1.6 \times {{10}^{ - 19}} \times 2 \times {{10}^{ - 2}}}} \\\ \Rightarrow B = 0.05\,{\text{T}} \\\
Therefore, the magnitude of the magnetic field is found to be $0.05\,{\text{T}}$.

Now, to find the magnitude of the electric field:
E = v \times B \\\ \Rightarrow E = {10^5} \times 0.05 \\\ \Rightarrow E = 0.5 \times {10^4}\,{\text{N}}\,{{\text{C}}^{ - 1}} \\\
Hence, the magnitude of the electric field is found to be $0.5 \times {10^4}\,{\text{N}}\,{{\text{C}}^{ - 1}}$.

Note: While solving the problem, most of the students have a confusion regarding when the proton goes undeflected. It is the condition in which the magnitude of the electric field is equal to the magnetic field. In this condition, both the fields cancel out each other. It is important to remember that, higher is the strength of the electric field, higher is the velocity acquired by the particle.