question_answer

A straight wire of length \[({{\pi }^{2}})\] metre is carrying a current of 2A 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                                       [Haryana CEE 1996]

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 \[2\pi r=l\] Þ \[r=\frac{l}{2\pi }\] \[\vec{F}=i(\vec{L}\times \vec{B})\] 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 \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2\pi i}{r}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2\pi \times 2}{\pi /2}\]Þ \[\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{1}{50}\]