An electron revolves in a circle of radius $0.4 Ã…$ with a speed of $10^6 \mathrm{~m} / \mathrm{s}$ in a…

An electron revolves in a circle of radius $0.4 Ã…$ with a speed of $10^6 \mathrm{~m} / \mathrm{s}$ in a hydrogen atom. The magnetic field produced at the centre of the orbit due to the motion of the electron (in Tesla) is : $\left[\mu_0=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right.$, charge on the electron $\left.=1.6 \times 10^{-19} \mathrm{C}\right]$
  1. $0.1$
  2. $1.0$
  3. $10$
  4. $100$

Solution

$ r=0.4 Ã…,=0.4 \times 10^{-10} \mathrm{~m}, v=10^6 \mathrm{~m} / \mathrm{s} $ The moving electron will constitute an electric current Electric current $i=\frac{q}{t}$ $ \begin{aligned} i & =\frac{q}{\frac{2 \pi r}{v}}=\frac{q v}{2 \pi r} \\ i & =\frac{1.6 \times 10^{-19} \times 10^6}{2 \pi \times 0.4 \times 10^{-10}} \\ & =\frac{1.6 \times 10^{-3}}{0.8 \pi} \end{aligned} $ Magnetic produced at the centre $ \begin{aligned} B & =\frac{\mu_0 i}{2 r} \\ B & =\frac{4 \pi \times 10^{-7}}{2 \times 0.4 \times 10^{-10}} \times \frac{1.6 \times 10^{-3}}{0.8 \pi} \\ & =\frac{4 \pi \times 1.6}{0.8 \times 0.8 \pi}=\frac{6.4}{0.64} \\ & =10 \text { tesla } \end{aligned} $

Asked in: AP EAMCET 2002

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