question_answer

Two long parallel wires are at a distance of 1 m. Both of them carry one ampere of current. The force of attraction per unit length between the two wires is:                                                                                       [AIPMT 1998]

A)  \[2\times {{10}^{-7}}\,N/m\]   

B)  \[2\times {{10}^{-8}}\,N/m\] C)  \[5\times {{10}^{-8}}\,N/m\] D)       \[{{10}^{-7}}\,N/m\] Correct Answer: A Solution : [a] Magnetic force between parallel wires per unit length is

\[\frac{F}{l}=\frac{{{\mu }_{0}}}{2\pi }\times \frac{{{i}_{1}}\,\,{{i}_{2}}}{r}\]

where \[{{i}_{1}}\] and \[{{i}_{2}}\] are the currents in wires 1 and 2 respectively and r is the distance between them. Since, it is given that between two wires, there is a force of attraction, so the direction of currents in both will be the same.

Here,     \[{{i}_{1}}={{i}_{2}}=1\,A,\,r=1\,m,\]

\[{{\mu }_{0}}=4\pi \times {{10}^{-7}}\,T-m/A\]

\[\therefore \]      \[\frac{F}{l}=\frac{4\pi \times {{10}^{-7}}}{2\pi }\times \frac{1\times 1}{1}\]

\[=2\times {{10}^{-7}}\,N/m\]