question_answer

A straight wire of length \[\left( {{\pi }^{2}} \right)\] 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

A) \[50 : 1\]

B)               \[1 : 50\]

C) \[100 : 1\]                     

D) \[1 : 100\]

Correct Answer: B

Solution :

If a wire of length l is bent in the form of a circle of radius r, then \[2\pi r=1\,\,\Rightarrow \,\,r=\frac{l}{2\pi }\,\] \[r=\frac{l}{2\pi }=\frac{{{\pi }^{2}}}{2\pi }=\frac{\pi }{2}\] Magnetic field due to straight wire \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2i}{r}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\times 2}{1\times {{10}^{-2}}}\] also magnetic field due to circular loop \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2\pi i}{r}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2\pi \times 2}{\pi /2}\] \[\Rightarrow \,\,\,\,\,\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{1}{50}\]