Magnetic field induction at the centre of a circular coil of radius $5 \mathrm{~cm}$ and carrying a current $0.9 \mathrm{~A}$ is (in SI units) $\left(\varepsilon_0= ight.$ absolute permittivity of air in SI units, velocity of light $=3 imes 10^8 \mathrm{~ms}^{-1}$ )

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Magnetic field induction at the centre of a circular coil of radius $5 \mathrm{~cm}$ and carrying a current $0.9 \mathrm{~A}$ is (in SI units) $\left(\varepsilon_0=ight.$ absolute permittivity of air in SI units, velocity of light $=3 	imes 10^8 \mathrm{~ms}^{-1}$ )

Aicharya · Accepted Answer

Magnetic field induction at the centre of circular coil,

$\begin{aligned} & c^2=\frac{1}{\mu_0 \varepsilon_0} \\ & \therefore \quad \mu_0=\frac{1}{\varepsilon_0 c^2} \\ & \end{aligned}$ So, from Eq. (i), we get $\begin{aligned} B & =\frac{1}{\varepsilon_0 c^2} \frac{I}{2 r} \\ & =\frac{1}{\varepsilon_0 \times\left(3 \times 10^8\right)^2} \times \frac{0.9}{2 \times 5 \times 10^{-2}} \\ & =\frac{1 \times 9}{\varepsilon_0 \times 9 \times 10^{16}}=\frac{1}{\varepsilon_0 \times 10^{16}} \end{aligned}$

Magnetic field induction at the centre of a circular coil of radius $5 \mathrm{~cm}$ and carrying a current…

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