Question

An electron revolves in a circle of radius $0.4 \mathring{A}$ with a speed of $10^5 ms^{-1}.$ The magnitude of the magnetic field, produced at the centre of the circular path due to the motion of the electron, in $Wb \, \, m^{-2}$ is

• A. 0.01
• B. 10
• C. 1
• D. 0.005

Answer 1

Answer: (C)

1

Answer 2

Magnetic field at the centre
\hspace20mm B= \frac {\mu _0qv}{4 \pi r^2}
Given $\frac {\mu_0}{4 \pi}=10^{-7}, q=1.6 \times 10^{-19}C,$
\hspace20mm v=10^5ms^{-1}
\hspace20mm r=0.4 \mathring{A}=0.4 \times 10^{-10}m
\therefore \hspace20mm B=10^{-7}\times \frac {1.6 \times 10^{-19}\times 10^5}{(0.4 \times 10^{-10})^2}=1 \, Wb \, m^{-2}