A particle having a charge 50 e is revolving in a circular path of radius 0.4 m with 1 r.p.s. The magnetic…

A particle having a charge 50 e is revolving in a circular path of radius 0.4 m with 1 r.p.s. The magnetic field produced at the centre of the circle is $\left(\mu_0=4 \pi \times 10^{-7}\right.$ SI units and $\left.\mathrm{e}=1 \cdot 6 \times 10^{-19} \mathrm{c}\right)$
  1. $10^{-7} \mu_0$
  2. $10^{-10} \mu_0$
  3. $10^{-14} \mu_0$
  4. $10^{-17} \mu_0$

Solution

$\begin{aligned} B & =\frac{\mu_0 \mathrm{qf}}{2 \mathrm{r}}=\frac{\mu_0 \times 50}{2 \times 0.4} \\ & =\frac{\mu_0 \times 50 \times 1.6 \times 10^{-9}}{0.8} \quad \ldots(\because \mathrm{f}=1 \text { r.p.s. }) \\ & =10^{-17} \mu_0\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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