The magnetic field at the centre O of the arc shown in the figure is

2 $I (\sqrt{2} + π) \times \frac{10^{- 7}}{r}$

2 $I (\sqrt{2} + \frac{π}{4}) \times \frac{10^{- 7}}{r}$

$I (\sqrt{2} + π) \times \frac{10^{- 7}}{r}$

$I (\sqrt{2} + \frac{π}{4}) \times \frac{10^{- 7}}{r}$

Solution

2 $I (\sqrt{2} + \frac{π}{4}) \times \frac{10^{- 7}}{r}$

Here, a = $\frac{r}{\sqrt{2}}$

Magnetic field at point O due to AB is

B₁ = $\frac{μ_{0}}{4 π} \frac{I}{a}$

= $\frac{μ_{0} I}{4 π (\frac{r}{\sqrt{2}})}$

Magnetic field at point O due to BCD is

B₂ = $\frac{μ_{0} I}{4 π a} (\frac{π}{2})$

Magnetic fieid at point O due to DE is

B₃ = $\frac{μ_{0} I}{4 π α}$

B₃ = $\frac{μ_{0} I}{4 π (\frac{r}{\sqrt{2}})}$

Resultant magnetic field at point O is

B = B₁ + B₂+ B₃

= $\frac{μ_{0} I}{4 π (\frac{r}{\sqrt{2}})} + \frac{μ_{0} I}{4 πr} (\frac{π}{2}) + \frac{μ_{0} I}{4 π (\frac{r}{\sqrt{2}})}$

= $\frac{μ_{0} I}{4 πr} (\sqrt{2} + \frac{π}{2} + \sqrt{2})$

B = $\frac{μ_{0} I}{4 πr} (2 \sqrt{2} + \frac{π}{2})$

= $\frac{μ_{0} 2 I}{4 πr} (\sqrt{2} + \frac{π}{4})$

B = $\frac{10^{- 7} \times 2 I}{r} (\sqrt{2} + \frac{π}{4})$ .... $[∵ \frac{μ_{0}}{4 π} = 10^{- 7}]$

Some More Questions From Moving Charges And Magnetism Chapter

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