A) \[4 : 3\]
B) \[3 : 1\]
C) \[5:3\]
D) \[2:3\]
Correct Answer: C
Solution :
The wires are in parallel and ratio of their resistances are\[3:4:5\], Hence currents in wires are in the ratio \[\frac{1}{3}:\frac{1}{4}:\frac{1}{5}\] \[{{i}_{1}}=\frac{k}{3}\] \[{{i}_{2}}=\frac{k}{4}\] \[{{i}_{3}}=\frac{k}{5}\] Force between top and middle wire \[{{F}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \,\,\frac{2{{i}_{1}}{{i}_{2}}}{{{r}_{1}}}\,\,=\,\,\frac{{{\mu }_{0}}}{4\pi }\,\,\times \,\,\frac{2\left( \frac{1}{3} \right)\left( \frac{1}{4} \right){{k}^{2}}}{{{r}_{1}}}\,\] Force between bottom and middle wire \[{{F}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\frac{\left( \frac{1}{4} \right)\left( \frac{1}{5} \right){{k}^{2}}}{{{r}_{2}}}\] As the forces are equal and opposite so \[{{F}_{1}}={{F}_{2}}\,\,\Rightarrow \,\,\frac{{{r}_{1}}}{{{r}_{2}}}\,=\,\frac{5}{3}\]You need to login to perform this action.
You will be redirected in
3 sec