- AZero, since charge on electron in constant
- B$\frac{\mu_0\text{er}^2}{4\text{R}}\alpha$
- C$\frac{\mu_0\text{er}^2}{4\pi\text{R}}\alpha$
- DNone of these
Explanation:
Using the formulas-
$\text{I}=\frac{\text{e}}{\text{T}}=\frac{\text{e}\omega}{2\pi}$
B at center $=\frac{\mu_0}{2\text{R}}=\frac{\mu_0\text{e}\omega}{4\pi\text{R}}$
$\phi=\text{B}.\pi\text{r}^2=\frac{\mu_0}{4\pi\text{R}}\text{e}\omega\pi\text{r}^2$
$=\frac{\mu_0\text{e}\omega\text{r}^2}{4\text{R}}$
$\text{e}=-\frac{\text{d}\phi}{\text{dt}}=\frac{\mu\text{e}\text{r}^2\alpha}{4\text{R}}$
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