A straight wire carrying current 'I' is bent into a semi-circular arc of radius ${ }^{\prime}…

A straight wire carrying current 'I' is bent into a semi-circular arc of radius ${ }^{\prime} \mathrm{r}^{\prime}$, as shown. The magnitude of magnetic field at point '0' due to semi-circular arc is $\left(\mu_{0}=\right.$ Permeability of free space)
  1. $\frac{\mu_{0} \text { I }}{4 \mathrm{r}}$
  2. $\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$
  3. $\frac{\mu_{0} \mathrm{I}}{\mathrm{r}^{2}}$
  4. $\frac{\mu_{0} \mathrm{I}}{\mathrm{r}}$

Solution

$B=\frac{\mu_{0} I}{4 r}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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