A) \[\frac{{{\mu }_{0}}I}{4\pi r}\]
B) \[\frac{{{\mu }_{0}}I}{4\pi r}+\frac{{{\mu }_{0}}I}{2\pi r}\]
C) \[\frac{{{\mu }_{0}}I}{4r}+\frac{{{\mu }_{0}}I}{4\pi r}\]
D) \[\frac{{{\mu }_{0}}I}{4r}-\frac{{{\mu }_{0}}I}{4\pi r}\]
Correct Answer: C
Solution :
Magnetic field due to different parts are B1 = 0 \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{\pi i}{r}\]¤ \[{{B}_{3}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{i}{r}\]¤ \[\therefore \ {{B}_{net}}={{B}_{2}}+{{B}_{3}}=\frac{{{\mu }_{0}}i}{4r}+\frac{{{\mu }_{0}}i}{4\pi r}\]You need to login to perform this action.
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