question_answer

A long straight wire carrying a current of 30 A is placed in an external uniform magnetic field of induction\[4\times {{10}^{-4}}T\]. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point 2.0 cm away from the wire is\[({{\mu }_{0}}=4\pi \times {{10}^{-7}}H{{m}^{-1}})\]

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 \[B=\sqrt{B_{1}^{2}+B_{2}^{2}}\] \[=(\sqrt{{{(4)}^{2}}+{{(3)}^{2}}})\times {{10}^{-4}}\] \[B=5\times {{10}^{-4}}T\]