A) \[{{10}^{-4}}\]
B) \[3\times {{10}^{-4}}\]
C) \[5\times {{10}^{-4}}\]
D) \[6\times {{10}^{-4}}\]
Correct Answer: C
Solution :
[c] Magnetic field due to wire \[B=\frac{{{\mu }_{0}}I}{2\pi r}=\frac{4\pi \times {{10}^{-7}}}{2\pi }\times \frac{30}{2\times {{10}^{-2}}}\] \[=3\times {{10}^{-4}}T\] This magnetic field will be perpendicular to external magnetic field. \[\therefore \] Net magnetic field \[B=\sqrt{{{B}^{2}}+B_{0}^{2}}\] \[=\sqrt{{{(3\times {{10}^{-4}})}^{2}}+{{(4\times {{10}^{-4}})}^{2}}}\] \[=5\times {{10}^{-4}}T\]You need to login to perform this action.
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