question_answer

Two parallel wires 1 m apart carry currents of 1 A and 3 A respectively in opposite directions. The force per unit length acting between these two wires is

A)  \[6\times {{10}^{-7}}N\text{ }{{m}^{-1}}\text{ }attractive\]

B)  \[6\times {{10}^{-5}}\text{ }N\text{ }{{m}^{-1}}\text{ }attractive\]

C)  \[6\times {{10}^{-7}}\text{ }N\text{ }{{m}^{-1}}\text{ }repulsive\]

D)  \[6\times {{10}^{-5}}\text{ }N\text{ }{{m}^{-1}}\text{ }repulsive\]

Correct Answer: C

Solution :

: The force per unit length acting between two parallel wires carrying currents \[{{I}_{1}}\] and \[{{I}_{2}}\]and placed a distance d apart is \[f=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi d}\] Here,  \[{{I}_{1}}=1A,\,{{I}_{2}}=3A,\,\,d=1m\] \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm\,{{A}^{-1}}\] \[\therefore \]  \[f=\frac{(4\pi \times {{10}^{-7}}T\,m\,{{A}^{-1}})(1A)(3A)}{2\pi (1m)}\] \[=6\times {{10}^{-7}}\,N\,{{m}^{-1}}\] As the currents are in opposite directions, so f is repulsive.