The magnetic field due to current carrying circular loop of radius $6 \mathrm{~cm}$ at a point on the axis at a distance of $8 \mathrm{~cm}$ from centre is $27 \mu \mathrm{T}$. The magnetic field at the centre of the current carrying loop is.

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Aicharya · Accepted Answer

( Magnetic field due to a current carrying coil of radius $R$ on its axis at a distance $x$ from its centre is $ B_{\text {axis }}=\frac{\mu_0 N I R^2}{2\left(x^2+R^2\right)^{\frac{3}{2}}} $

Here, $x=8 \mathrm{~cm},=8 \times 10^{-2} \mathrm{~m}$ $ R=6 \mathrm{~cm}=6 \times 10^{-2} \mathrm{~m} $ and $B=27 \mu \mathrm{T}=27 \times 10^{-6} \mathrm{~T}$ Substituting these values in eq. (i), we get $ \begin{aligned} 27 \times 10^{-6} & =\frac{\mu_0 N I \times\left(6 \times 10^{-2}\right)^2}{2\left[\left(8 \times 10^{-2}\right)^2+\left(6 \times 10^{-2}\right)^2\right]^{3 / 2}} \\ & =\frac{\mu_0 N I \times\left(6 \times 10^{-2}\right)^2}{\left.264 \times 10^{-4}+36 \times 10^{-4}\right]^{3 / 2}} \\ \Rightarrow 27 \times 10^{-6} & =\frac{\mu N I \times 36 \times 10^{-4}}{2 \times\left(10^{-2}\right)^{3 / 2}} \\ \Rightarrow 27 \times 10^{-6} & =\frac{\mu_0 N I \times 36 \times 10^{-4}}{2 \times 10^{-3}} \\ \Rightarrow \quad \mu_0 N I & =\frac{27 \times 10^{-6} \times 2 \times 10^{-3}}{36 \times 10^{-4}} \\ & =\frac{3}{2} \times 10^{-5} \text { units } \end{aligned} $ Now, magnetic field at centre, $B_{\text {centre }}=\frac{\mu_0 N I}{2 R}$ Substituting values in above expression from eq. (ii) we get, $ \begin{aligned} B_{\text {centre }} & =\frac{\frac{3}{2} \times 10^{-5}}{2 \times 6 \times 10^{-2}}=\frac{1}{8} \times 10^{-3} \mathrm{~T} \\ & =\frac{1000}{8} \times 10^{-6} \mathrm{~T} \end{aligned} $ or $ B_{\text {centre }}=125 \mu \mathrm{T} $

The magnetic field due to current carrying circular loop of radius $6 \mathrm{~cm}$ at a point on the axis…

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