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 \[\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{1}{50}\]You need to login to perform this action.
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