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(B) For a cylindrical conductor of radius $R$, the magnetic field inside at a distance $r < R$ is given by $B_{in} = \frac{\mu_0 i r}{2 \pi R^2}$.Give

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$\frac{8}{3} \ T$

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(B) For a cylindrical conductor of radius $R$, the magnetic field inside at a distance $r Given that the point is $R/4$ inside from the surface, the distance from the center is $r = R - R/4 = 3R/4$.Thus, $B_{in} = \frac{\mu_0 i (3R/4)}{2 \pi R^2} = \frac{3 \mu_0 i}{8 \pi R} = 10 \ T$.From this, we find $\frac{\mu_0 i}{2 \pi R} = \frac{10 	imes 4}{3} = \frac{40}{3} \ T$.For a point outside at a distance $4R$ from the surface, the distance from the center is $r' = R + 4R = 5R$.The magnetic field outside is $B_{out} = \frac{\mu_0 i}{2 \pi r'} = \frac{\mu_0 i}{2 \pi (5R)} = \frac{1}{5} \left( \frac{\mu_0 i}{2 \pi R} ight)$.Substituting the value, $B_{out} = \frac{1}{5} 	imes \frac{40}{3} = \frac{8}{3} \ T$.

$A$ cylindrical conductor of radius $R$ carries a current $i$. The value of the magnetic field at a point, which is $R/4$ distance inside from the surface, is $10 \ T$. Find the value of the magnetic field at a point, which is $4R$ distance outside from the surface.

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