Question

An electron moving at the right angle to a uniform magnetic field completes a circular orbit in 1 microsecond. Find the magnetic field.

Answer 1

Hint: The charge in a magnetic field, moving with a velocity causes a force, which is called Lorentz force. The charged particle to complete a circular motion, it needs a centripetal force as well. So, you should identify the forces properly in this problem.

Formula Used:
The Lorentz force $\vec F$ is defined as
$\vec F = q\left( {\vec v \times \vec B} \right)$
where, $q$ is the charge of the particle, $\vec v$ is the velocity of the particle and $\vec B$ is the uniform magnetic field.
If a particle with mass $m$ moves with a velocity $v$ through a circular path with radius $r$ then the centripetal force is defined to be
$F = \dfrac{{m{v^2}}}{r}$

Complete step by step answer:
Given:
The electron completes a circular orbit in 1 microsecond.
To get: The magnetic field.
Step 1:
Let the electron moves with a velocity $v$ in the uniform field $B$
The electron moves at a right angle to the uniform magnetic field $B$.
Hence calculate the Lorentz force from eq (1)
$
  F = qvB\sin \theta \\
   \Rightarrow F = qvB\sin {90^ \circ } \\
   \Rightarrow F = qvB \\
 $
Step 2:
Let the electron moving along a circular path of radius $r$.
Let the mass of the electron is m.
Calculate the centripetal force to keep it in a circular orbit.
$F = \dfrac{{m{v^2}}}{r}$

Step 3:
Now you can see that the Lorentz force supplies the required centripetal force to keep it in a circular motion.
Now find the expression of the magnetic field
$
  F = qvB = \dfrac{{m{v^2}}}{r} \\
   \Rightarrow B = \dfrac{{mv}}{{qr}} \\
 $
Step 4:
By the problem, the electron completes one circular orbit.
So, you can calculate the angular velocity
$\omega = \dfrac{{2\pi }}{{1 \times {{10}^{ - 6}}}}rad.{s^{ - 1}}$

Step 5:
The mass of the electron is $m = 9.11 \times {10^{ - 31}}kg$ and the charge is $q = 1.6 \times {10^{ - 19}}C$.
By definition, we know that the angular velocity can be represented as
$\omega = \dfrac{v}{r}$
Hence, now put the values together to get the value of the magnetic field
$
B = \dfrac{{mv}}{{qr}} \\
\Rightarrow B = \dfrac{m}{q}\dfrac{v}{r} \\
\Rightarrow B = \dfrac{m}{q}\omega \\
\Rightarrow B = \dfrac{{9.11 \times {{10}^{ - 31}}}}{{1.6 \times {{10}^{ - 19}}}} \times \dfrac{{2\pi }}{{1 \times {{10}^{ - 6}}}}T \\
\Rightarrow B = \dfrac{{9.11 \times 2 \times 3.14 \times {{10}^{ - 31}}}}{{1.6 \times {{10}^{ - 19}} \times {{10}^{ - 6}}}}T = 3.53 \times {10^{ - 5}}T \\
\therefore B = 3.53 \times {10^{ - 5}}T \\
 $

If an electron moving at right angle to a uniform magnetic field completes a circular orbit in microsecond, then the magnetic field is $3.53 \times {10^{ - 5}}T$.

Note:
The Lorentz force is to be completely equal to the centripetal force as there are no other sources in the system. When the electron completes a circular orbit, that means it traverses $2\pi rad$ angular distance. You need to carefully notice that this angular distance is traversed in the said $1$ microsecond time interval.