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Question

(B) The magnetic field inside a cylindrical conductor at a distance $r$ from the axis is given by $B_{in} = \frac{\mu_0}{4\pi} \cdot \frac{2ir}{R^2}$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The magnetic field inside a cylindrical conductor at a distance $r$ from the axis is given by $B_{in} = \frac{\mu_0}{4\pi} \cdot \frac{2ir}{R^2}$.Given the point is $R/4$ inside from the surface,the distance from the axis is $r = R - R/4 = 3R/4$.Thus,$B_{in} = \frac{\mu_0}{4\pi} \cdot \frac{2i(3R/4)}{R^2} = 10 \ T$.The magnetic field outside the cylinder at a distance $r'$ from the axis is $B_{out} = \frac{\mu_0}{4\pi} \cdot \frac{2i}{r'}$.Given the point is $4R$ outside from the surface,the distance from the axis is $r' = R + 4R = 5R$.Taking the ratio: $\frac{B_{in}}{B_{out}} = \frac{(3R/4) / R^2}{1 / (5R)} = \frac{3R}{4R^2} \cdot 5R = \frac{15}{4}$.Therefore,$B_{out} = B_{in} \cdot \frac{4}{15} = 10 \cdot \frac{4}{15} = \frac{40}{15} = \frac{8}{3} \ T$.

$A$ 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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