NEET AIPMT SOLVED PAPER 1998

  • 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:                                                                      

    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 :

                    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,                 \[{{i}_{1}}={{i}_{2}}=1\,A,\,r=1\,m,\]                 \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}\,T-m/A\]                 \[\therefore \]  \[\frac{F}{l}=\frac{4\pi \times {{10}^{-7}}}{2\pi }\times \frac{1\times 1}{1}\]                 \[=2\times {{10}^{-7}}\,N/m\]


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