question_answer

                                Two wires are held perpendicular to the plane of paper at 5 m apart. They carry currents of 2.5 A and 5 A in same direction. Then the magnetic field strength (b) at a point midway between the wires will be:

A)                                                                                                                                                                                         \[\frac{{{\mu }_{0}}}{4\pi }T\]   

B)                 \[\frac{{{\mu }_{0}}}{2\pi }T\]

C)                 \[\frac{3{{\mu }_{0}}}{2\pi }T\] 

D)                 \[\frac{3{{\mu }_{0}}}{4\pi }T\]

Correct Answer: B

Solution :

                                Key Idea: At mid-paint the directions of magnetic field due to both wires carrying current in same direction are opposite.                 According to Maxwell's right handed screw rule, the magnetic field at right hand of wire 1 is perpendicular to die paper going inwards shown by \[\otimes \]. Similarly, die magnetic field at left hand of wire 2 is perpendicular to paper coming out shown by \[\odot \]. Thus, die two fields are opposite to each other.                 Therefore, net magnetic field                 \[B={{B}_{1}}-{{B}_{2}}\]                 \[=\frac{{{\mu }_{0}}{{i}_{1}}}{2\pi {{r}_{1}}}-\frac{{{\mu }_{0}}{{i}_{2}}}{2\pi {{r}_{2}}}\]                 At mid point \[{{r}_{1}}={{r}_{2}}=r=\frac{5}{2}\,=2.5\,cm\]                 Hence,  \[B=\frac{{{\mu }_{0}}}{2\pi }\left( \frac{{{i}_{1}}}{r}-\frac{{{i}_{2}}}{r} \right)\]                 \[=\frac{{{\mu }_{0}}}{2\pi }\left( \frac{5}{2.5}-\frac{2.5}{2.5} \right)\]                 \[=\frac{{{\mu }_{0}}}{2\pi }(2-1)\]                 \[=\frac{{{\mu }_{0}}}{2\pi }T\]