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                 \[{{B}_{I}}=\frac{{{\mu }_{o}}}{4\pi }\frac{2{{i}_{1}}}{x}\otimes \] and        \[{{B}_{II}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{i}_{2}}}{x}\varnothing \] So, the net magnetic field at\[O\]                 \[{{B}_{net}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{x}\times ({{i}_{1}}-{{i}_{2}})\] \[\Rightarrow \]               \[10\times {{10}^{-6}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2}{x}({{i}_{1}}-{{i}_{2}})\]      ... (i) when the direction of \[{{i}_{2}}\] is reversed                 \[{{B}_{I}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2{{i}_{1}}}{x}\otimes \] and        \[{{B}_{II}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2{{i}_{2}}}{x}\otimes \] So, net magnetic field at \[O\]                 \[{{B}_{net}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{x}\times ({{i}_{1}}+{{i}_{2}})\] \[\Rightarrow \]               \[40\times {{10}^{-6}}=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2}{x}({{i}_{1}}+{{i}_{2}})\]     ... (ii) Dividing Eq. (ii) by Eq. (i), we get                 \[\frac{{{i}_{1}}+{{i}_{2}}}{{{i}_{1}}-{{i}_{2}}}=\frac{4}{1}\] \[\Rightarrow \]               \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{5}{3}\]