A) \[{{10}^{-4}}\]
B) \[3\times {{10}^{-4}}\]
C) \[5\times {{10}^{-4}}\]
D) \[6\times {{10}^{-4}}\]
Correct Answer: C
Solution :
Magnetic field at a point 2.0 cm from the straight conductor carrying current of 3QA is \[B=\frac{{{\mu }_{0}}i}{2\pi R}=\frac{4\pi \times {{10}^{-7}}\times 30}{2\pi \times 2\times {{10}^{-2}}}=3\times {{10}^{-4}}T\] The conductor is placed in the external field \[{{\vec{B}}_{0}}\]which is directed perpendicular to\[\vec{B}.\]. Therefore, the resultant magnetic field is \[{{B}_{r}}=\sqrt{{{B}^{2}}+B_{0}^{2}}+\sqrt{{{3}^{2}}+{{4}^{2}}}\times {{10}^{-4}}T\] \[=5\times {{10}^{-4}}T\] Hence, the correction option is [c].
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