Question

Two long parallel wires carry currents $ {{i}_{1}} $ and $ {{i}_{2}} $ such that $ {{i}_{1}}>{{i}_{2}} $ . When the currents are in the same direction, the magnetic field at a point midway between the wires is $ 6\times {{10}^{-6}}T $ . If the direction of $ {{i}_{2}} $ is reversed, the field becomes $ 3\times {{10}^{-5}}T. $ The ratio $ \frac{{{i}_{1}}}{{{i}_{2}}} $ is

• A. $ \frac{1}{2} $
• B. 2
• C. $ \frac{2}{3} $
• D. $ \frac{3}{2} $
• E. $ \frac{1}{5} $

Answer 1

Answer: (D)

When the currents are in the same direction then magnetic field
$ B=\frac{{{\mu }_{0}}}{4\pi }\times \frac{1}{d}[{{i}_{}}_{1}-{{i}_{2}}] $
$ 6\times {{10}^{-6}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{d}[{{i}_{1}}-{{i}_{2}}] $ ...(i)
When the currents are in the reversed direction then magnetic field
$ {{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{d}[{{i}_{1}}-(-{{i}_{2}})] $
Or $ {{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{d}[{{i}_{1}}+{{i}_{2}}] $
or $ 3\times {{10}^{-5}}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2}{d}[{{i}_{1}}+{{i}_{2}}] $ ...(ii) Dividing Eq. (i) by Eq. (ii)
$ \frac{{{i}_{1}}-{{i}_{2}}}{{{i}_{1}}+{{i}_{2}}}=\frac{6\times {{10}^{-6}}}{3\times {{10}^{-5}}} $
Or $ \frac{{{i}_{1}}-{{i}_{2}}}{{{i}_{1}}+{{i}_{2}}}=\frac{2}{10} $
Or $ 5{{i}_{1}}-5{{i}_{2}}={{i}_{1}}+{{i}_{2}} $
Or $ 4{{i}_{1}}=6{{i}_{2}} $
Or $ \frac{{{i}_{1}}}{{{i}_{2}}}=\frac{6}{4} $
Or $ \frac{{{i}_{1}}}{{{i}_{2}}}=\frac{3}{2} $