A) \[{{10}^{-4}}\]
B) \[3\times {{10}^{-4}}\]
C) \[5\times {{10}^{-4}}\]
D) \[6\times {{10}^{-4}}\]
Correct Answer: C
Solution :
Given, \[i=30A,{{B}_{1}}=4\times {{10}^{-4}}T,\] \[r=2\text{ }cm=0.02\text{ }m\] Magnetic field induction at point P due to current carrying wire is \[{{B}_{2}}=\frac{{{\mu }_{0}}i}{2\pi r}=\frac{4\pi \times {{10}^{-7}}\times 30}{2\pi \times 0.02}\] \[=3\times {{10}^{-4}}T\] The direction of\[{{B}_{2}}\]will be perpendicular to\[{{B}_{1}},\]then the magnitude of the resultant magnetic induction \[B=\sqrt{B_{1}^{2}+B_{2}^{2}}\] \[=(\sqrt{{{(4)}^{2}}+{{(3)}^{2}}})\times {{10}^{-4}}\] \[B=5\times {{10}^{-4}}T\]You need to login to perform this action.
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