MHT CET · Physics · Magnetic Effects of Current
An electron makes a full rotation in a circle of radius \(0.8 \mathrm{~m}\) in one second. The magnetic field at the centre of the circle is
\(
\left(\mu_0=4 \pi \times 10^{-7} \text { SI units }\right)
\)
- A \(4 \pi \times 10^{-26} \mathrm{~T}\)
- B \(2 \pi \times 10^{-26} \mathrm{~T}\)
- C \(4 \pi \times 10^{-19} \mathrm{~T}\)
- D \(2 \pi \times 10^{-19} \mathrm{~T}\)
Answer & Solution
Correct Answer
(A) \(4 \pi \times 10^{-26} \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \omega=\frac{2 \pi}{T} \\
& I=\frac{q \omega}{2 \pi}=\frac{1.6 \times 10^{-19} \times 2 \pi}{2 \pi} \quad \ldots .\left(\because I=\frac{q}{t}\right) \\
& I=1.6 \times 10^{-19} A
\end{aligned}
\)
\(\therefore \quad\) The magnetic field at the centre of the circle is:
\(
\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19}}{2 \times 0.8} \\
& \mathrm{~B}=4 \pi \times 10^{-26} \mathrm{~T}
\end{aligned}
\)
\begin{aligned}
& \omega=\frac{2 \pi}{T} \\
& I=\frac{q \omega}{2 \pi}=\frac{1.6 \times 10^{-19} \times 2 \pi}{2 \pi} \quad \ldots .\left(\because I=\frac{q}{t}\right) \\
& I=1.6 \times 10^{-19} A
\end{aligned}
\)
\(\therefore \quad\) The magnetic field at the centre of the circle is:
\(
\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19}}{2 \times 0.8} \\
& \mathrm{~B}=4 \pi \times 10^{-26} \mathrm{~T}
\end{aligned}
\)
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