Solveeit Logo
Login

Question

Question: An electron is revolving in a circular path of radius 2.0 \(\times\) 10\(^{-10}\) m with a uniform s...

An electron is revolving in a circular path of radius 2.0 ×\times 10−10^{-10} m with a uniform speed of 3 ×\times 106^6 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

Explanation

Solution

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^{-10} m, an electron whose charge q is 1.6 ×\times 10−19^{-19} C is revolving in this circular path with a constant speed of v = 3 ×\times 106^6 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=qti = \dfrac{q}{t}
i=q2πrvi = \dfrac{q}{{\dfrac{{2\pi r}}{v}}} where t=distance/speed and distance is the circumference of the circular path i.e. 2πr
i=qv2πri = \dfrac{{qv}}{{2\pi r}}
The magnetic induction at the centre of the circular path,B=μ0i2rB = \dfrac{{{\mu _0}i}}{{2r}}
We substitute the value of current i in this formula
B=μ0(qv2πr)2r B=4π×10−7(qv2πr)2r B=4π×10−7×qv4πr2 B=qv×10−7r2 B=1.6×10−19×3×106×10−7(2×10−10)2=4.8×10−19−7+64×10−20 B=1.2×10−2010−20=1.2T 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 μ0μ_0 is called permeability of free space and its value is 4π×\times10−7^{-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.