Respuesta :
Answer:
The magnetic field at the center of flat coil is [tex]\dfrac{\mu_{0}I}{8a}[/tex].
Explanation:
Given that,
Radius [tex]r = a(1+\theta^2)[/tex]
We need to calculate the magnetic field at the center of flat coil
Using Biot-savart law
[tex]dB=\dfrac{\mu_{0}Idl\sin\alpha}{4\pi\times r^2}[/tex]
Here, [tex]\alpha =90^{/circ}[/tex]
[tex] dl=rd\theta[/tex]
Then, the magnetic field
[tex]dB=\dfrac{\mu_{0}Ird\theta\sin90}{4\pi\times r^2}[/tex]
[tex]dB=\dfrac{\mu_{0}Id\theta}{4\pi\times r}[/tex]
Put the value of r
[tex]dB=\dfrac{\mu_{0}Id\theta}{4\pi\times a(1+\theta^2)}[/tex]
[tex]dB=\dfrac{\mu_{0}I}{4\pi a}\int_{0}^{\infty}{\dfrac{d\theta}{(1+\theta^2)}}[/tex]
[tex]dB=\dfrac{\mu_{0}I}{4\pi a}(1(\tan^2\theta))_{0}^{\infty}[/tex]
[tex]dB=\dfrac{\mu_{0}I}{4\pi a}(\dfrac{\pi}{2}-0)[/tex]
[tex]dB=\dfrac{\mu_{0}I}{4\pi a}\times\dfrac{\pi}{2}[/tex]
[tex]dB=\dfrac{\mu_{0}I}{8a}[/tex]
Hence, The magnetic field at the center of flat coil is [tex]\dfrac{\mu_{0}I}{8a}[/tex].
