A long straight wire of radius a carries a steady current $I .$ The current is uniformly distributed over its cross-section. The ratio of the magnetic fields $B$ and $B',$ at radial distances $\frac{a}{2}$ and $2 a$ respectively, from the axis of the wire is:

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$\frac{1}{4}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

For points inside the wire i.e., $(\begin{array}{ll} r R \end{array})$

Magnetic field $B = \frac{μ_{0} I r}{2 π R^{2}}$

For points outside the wire $(\begin{array}{ll} r R \end{array})$

Magnetic field, $B^{'} = \frac{μ_{0} I}{2 π R}$

$∴ \frac{B}{B^{'}} = \frac{\frac{μ_{0} l (\frac{a}{2})}{2 π a^{2}}}{\frac{μ_{0} I}{2 π (2 a)}} = 1 : 1$

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