MHT CET · Physics · Magnetic Effects of Current
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 \(\left.\mathrm{e}=1 \cdot 6 \times 10^{-19} \mathrm{c}\right)\)
- A \(10^{-7} \mu_0\)
- B \(10^{-10} \mu_0\)
- C \(10^{-14} \mu_0\)
- D \(10^{-17} \mu_0\)
Answer & Solution
Correct Answer
(D) \(10^{-17} \mu_0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} B & =\frac{\mu_0 \mathrm{qf}}{2 \mathrm{r}}=\frac{\mu_0 \times 50}{2 \times 0.4} \\ & =\frac{\mu_0 \times 50 \times 1.6 \times 10^{-9}}{0.8} \quad \ldots(\because \mathrm{f}=1 \text { r.p.s. }) \\ & =10^{-17} \mu_0\end{aligned}\)
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