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A cylindrical conductor of radius $R$ carries a current i. The value of magnetic field at a point which is $\frac{\mathrm{R}}{4}$ distance inside from the surface is $10 \mathrm{~T}$. The value of magnetic field at point which is $4 \mathrm{R}$ distance outside from the surface
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The correct answer is:
$\frac{8}{3} \mathrm{~T}$
Magnetic field inside the cyclindrical conductor $B_{\text {in }}=\frac{\mu_{0}}{4 \pi} \frac{2 i r}{R^{2}}$
$(R=$ radius of cylinder and $r=$ distance of observation point from axis of cylinder) Magnetic field outside the cylinder at a
distance $r^{\prime}$ from its axis, $B_{\text {out }}=\frac{\mu_{0}}{4 \pi} \frac{2 i}{r^{\prime}}$
$$
\begin{array}{l}
\Rightarrow \frac{B_{\mathrm{in}}}{B_{\mathrm{out}}}=\frac{r r^{\prime}}{R^{2}} \Rightarrow \frac{10}{B_{\mathrm{out}}}=\frac{\left(R-\frac{R}{4}\right)(R+4 R)}{R^{2}} \\
\Rightarrow B_{\mathrm{out}}=\frac{8}{3} T
\end{array}
$$
$(R=$ radius of cylinder and $r=$ distance of observation point from axis of cylinder) Magnetic field outside the cylinder at a
distance $r^{\prime}$ from its axis, $B_{\text {out }}=\frac{\mu_{0}}{4 \pi} \frac{2 i}{r^{\prime}}$
$$
\begin{array}{l}
\Rightarrow \frac{B_{\mathrm{in}}}{B_{\mathrm{out}}}=\frac{r r^{\prime}}{R^{2}} \Rightarrow \frac{10}{B_{\mathrm{out}}}=\frac{\left(R-\frac{R}{4}\right)(R+4 R)}{R^{2}} \\
\Rightarrow B_{\mathrm{out}}=\frac{8}{3} T
\end{array}
$$
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