A straight wire of length π² metre is carrying a current of 2 A and the magnetic field due to it is measured at a point distant 1 cm from it. If the wire is to be bent into a circle and is to carry the same current as before, the ratio of the magnetic field at its centre to that obtained in the first case would be approximately :

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1 : 50

50 : 1

1 : 100

100 : 1

answer is B.

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Detailed Solution

If a wire of length $ℓ$ is bent in the form of a circle of radius r then $2 π r = l \Rightarrow r = \frac{l}{2 π}$

$r = \frac{I}{2 π} = \frac{π^{2}}{2 π} = \frac{π}{2}$

Magnetic field due to straight wire

$B_{1} = \frac{μ_{0}}{4 π} \cdot \frac{2 i}{r} = \frac{μ_{0}}{4 π} \times \frac{2 \times 2}{∣ \times 10^{- 2}}$ also magnetic field due to circular loop

$\begin{array}{l} B_{2} = \frac{μ_{0}}{4 π} \cdot \frac{2 π i}{r} = \frac{μ_{0}}{4 π} \cdot \frac{2 π \times 2}{π / 2} \\ \Rightarrow \frac{B_{2}}{B_{1}} = \frac{1}{50} \end{array}$