question_answer

Two parallel long wires carry currents \[{{i}_{1}}\]and \[{{i}_{2}}\]with \[{{i}_{1}}>{{i}_{2}}.\]When the currents are in the same direction, the magnetic field midway between the wires is \[10\,\mu T.\] When the direction of \[{{i}_{2}}\] is reversed, it becomes \[40\mu T.\]Then, ratio of \[{{i}_{1}}/{{i}_{2}}\]is

A)  3 : 4

B)  5 : 3

C)  7 : 11

D)  11 : 7

Correct Answer: B

Solution :

 When the current in the wires is in same direction. Magnetic field at mid point O due to I and II wires are respectively \[{{B}_{I}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{i}_{1}}}{x}\otimes \] and \[{{B}_{II}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2{{i}_{2}}}{x}o/\] So, the net magnetic field at O \[{{B}_{net}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{x}\times ({{i}_{1}}-{{i}_{2}})\] \[\Rightarrow \] \[10\times {{10}^{-6}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2}{x}({{i}_{1}}-{{i}_{2}})\] ?(i) when the direction of \[{{i}_{2}}\]is reversed \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2{{i}_{1}}}{x}\otimes \] and \[{{B}_{II}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2{{i}_{2}}}{x}\otimes \] \[{{B}_{net}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{x}\times ({{i}_{1}}+{{i}_{2}})\] \[\Rightarrow \] \[40\times {{10}^{-6}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2}{x}({{i}_{1}}+{{i}_{2}})\] ?(ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{{{i}_{1}}+{{i}_{2}}}{{{i}_{1}}-{{i}_{2}}}=\frac{4}{1}\] \[\Rightarrow \] \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{5}{3}\]