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VITEEE 2010 Paper

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Question : 24 of 120

Marks: +1, -0

A cylindrical conductor of radius R carries a current i. The value of magnetic field at a point which is

\frac{R}{4}

distance inside from the surface is 10T. The value of magnetic field at point which is 4R distance outside from the surface

$\frac{4}{3}T$
$\frac{8}{3}T$
$\frac{40}{3}T$
$\frac{80}{3}T$

Solution:

Magnetic field inside the cylindrical conductor

B_{in}=\frac{\mu_{0}}{4\pi}\frac{2ir}{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' from its axis,

B_{out}=\frac{\mu_{0}}{4\pi}\frac{2i}{r'}

\Rightarrow\frac{B_{in}}{B_{out}}=\frac{rr'}{R^{2}}

\Rightarrow\frac{10}{B_{out}}=\frac{\left(R-\frac{R}{4}\right)(R+4R)}{R^{2}}

\Rightarrow B_{out}=\frac{8}{3}T

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