Question

An electron is revolving in a circular path of radius 2.0 $\times$ 10$^{-10}$ m with a uniform speed of 3 $\times$ 10$^6$ m/s. The magnetic induction at the centre of the circular path will be
A. 0.6T
B. 1.2T
C. O.12T
D. Zero

Answer 1

Hint: The current is calculated by dividing charge by time where time is defined as the distance /speed. The magnetic induction at the centre of the circular path is given by the formula μ0times the current divided by twice the radius.

Complete step by step answer:
In a circular path of radius R = 2.0 $\times$ 10$^{-10}$ m, an electron whose charge q is 1.6 $\times$ 10$^{-19}$ C is revolving in this circular path with a constant speed of v = 3 $\times$ 10$^6$ m/s. This orbiting electron behaves as a current loop with intensity of current i and it produces a magnetic field B as a body carrying charge (electron) when moving with a certain speed creates a magnetic field around it. The magnitude of magnetic field is directly proportional to the speed of the charge (electron). The number of magnetic lines of induction caused due to this field crossing per unit area normal to their direction is called the magnitude of magnetic induction (B).
The current is defined as the amount of charge passing through a conductor (due to electron) per unit time t i.e.,
$i = \dfrac{q}{t}$
$i = \dfrac{q}{{\dfrac{{2\pi r}}{v}}}$ where t=distance/speed and distance is the circumference of the circular path i.e. 2πr
$i = \dfrac{{qv}}{{2\pi r}}$
The magnetic induction at the centre of the circular path,$B = \dfrac{{{\mu _0}i}}{{2r}}$
We substitute the value of current i in this formula
$ B = \dfrac{{{\mu _0}\left( {\dfrac{{qv}}{{2\pi r}}} \right)}}{{2r}} \\
  B = \dfrac{{4\pi \times {{10}^{ - 7}}\left( {\dfrac{{qv}}{{2\pi r}}} \right)}}{{2r}} \\
  B = \dfrac{{4\pi \times {{10}^{ - 7}} \times qv}}{{4\pi {r^2}}} \\
  B = \dfrac{{qv \times {{10}^{ - 7}}}}{{{r^2}}} \\
  B = \dfrac{{1.6 \times {{10}^{ - 19}} \times 3 \times {{10}^6} \times {{10}^{ - 7}}}}{{{{\left( {2 \times {{10}^{ - 10}}} \right)}^2}}} = \dfrac{{4.8 \times {{10}^{ - 19 - 7 + 6}}}}{{4 \times {{10}^{ - 20}}}} \\
  B = 1.2 \times \dfrac{{{{10}^{ - 20}}}}{{{{10}^{ - 20}}}} = 1.2T \\
$
where $μ_0$ is called permeability of free space and its value is 4π$\times$10$^{-7}$

So, the correct answer is “Option B”.

Note:
The magnetic induction produced at the centre of the circular path is due to charge carrying particles i.e., electrons. The electron revolves in a circular path. So, the distance covered is equal to the circumference of the circle.