A) 2
B) 1/2
C) 1/4
D) 1
Correct Answer: D
Solution :
Magnetic field at centre due to smaller loop \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi {{i}_{1}}}{{{r}_{1}}}\] ..... (i) Due to Bigger loop \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi {{i}_{2}}}{{{r}_{2}}}\] So net magnetic field at centre \[B={{B}_{1}}-{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times 2\pi \left( \frac{{{i}_{1}}}{{{r}_{1}}}-\frac{{{i}_{2}}}{{{r}_{2}}} \right)\] According to question \[B=\frac{1}{2}\times {{B}_{1}}\] \[\Rightarrow \frac{{{\mu }_{0}}}{4\pi }.2\pi \left( \frac{{{i}_{1}}}{{{r}_{1}}}-\frac{{{i}_{2}}}{{{r}_{2}}} \right)=\frac{1}{2}\times \frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi {{i}_{1}}}{{{r}_{1}}}\] \[\frac{{{i}_{1}}}{{{r}_{1}}}-\frac{{{i}_{2}}}{{{r}_{2}}}=\frac{{{i}_{1}}}{2{{r}_{1}}}\Rightarrow \frac{{{i}_{1}}}{2{{r}_{1}}}=\frac{{{i}_{2}}}{{{r}_{2}}}\Rightarrow \frac{{{i}_{1}}}{{{i}_{2}}}=1\] \[\{{{r}_{2}}=2{{r}_{1}}\}\]
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