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 :
[b] 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. 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, \[B=\frac{{{\mu }_{0}}}{2\pi }\left( \frac{{{i}_{1}}}{r}-\frac{{{i}_{2}}}{r} \right)\] |
\[=\frac{{{\mu }_{0}}}{2\pi }\left( \frac{5}{2.5}-\frac{2.5}{2.5} \right)\] |
\[=\frac{{{\mu }_{0}}}{2\pi }(2-1)\] |
\[=\frac{{{\mu }_{0}}}{2\pi }T\] |
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