MHT CET · Physics · Magnetic Effects of Current
A particle carrying a charge equal to 100 times the charge on an electron is rotating one rotation per second in a circular path of radius 0.8 m . The value of magnetic field produced at the centre will be ( \(\mu_0=\) permeability of vacuum)
- A \(\frac{10^{-7}}{\mu_0}\)
- B \(10^{-17} \mu_0\)
- C \(10^{-6} \mu_0\)
- D \(10^{-7} \mu_0\)
Answer & Solution
Correct Answer
(B) \(10^{-17} \mu_0\)
Step-by-step Solution
Detailed explanation
Charged particle moving in a circular path acts like a current carrying coil.
Let I be the current due to the particle current is,
\(I=\frac{q}{t}=\frac{100 \times e}{1}=100 e\)
Magnetic field at the centre due to current carrying coil,
\(\begin{aligned}
\mathrm{B}_{\text {centre }} & =\frac{\mu_0}{4 \pi} \frac{2 \pi \mathrm{I}}{\mathrm{r}}=\frac{\mu_0}{4 \pi} \frac{2 \pi \times 100 \mathrm{e}}{\mathrm{r}} \\
& =\frac{\mu_0}{4} \frac{2 \times 100\left(1.6 \times 10^{-19}\right)}{0.8}=10^{-17} \mu_0
\end{aligned}\)
Let I be the current due to the particle current is,
\(I=\frac{q}{t}=\frac{100 \times e}{1}=100 e\)
Magnetic field at the centre due to current carrying coil,
\(\begin{aligned}
\mathrm{B}_{\text {centre }} & =\frac{\mu_0}{4 \pi} \frac{2 \pi \mathrm{I}}{\mathrm{r}}=\frac{\mu_0}{4 \pi} \frac{2 \pi \times 100 \mathrm{e}}{\mathrm{r}} \\
& =\frac{\mu_0}{4} \frac{2 \times 100\left(1.6 \times 10^{-19}\right)}{0.8}=10^{-17} \mu_0
\end{aligned}\)
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