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 tesia at a point 2.0 cm away from the wire is \[({{\mu }_{0}}=4\pi \times {{10}^{-7}}H/m)\]

A)  \[{{10}^{-4}}\]

B)  \[3\times {{10}^{-4}}\]

C)  \[5\times {{10}^{-4}}\]

D)  \[6\times {{10}^{-4}}\]

Correct Answer: C

Solution :

  Magnetic field induction at point P due to current carrying wire is \[{{B}_{2}}=\frac{{{\mu }_{0}}i}{2\pi r}\frac{4\pi \times {{10}^{-7}}\times 30}{2\pi \times 0.02}\] \[=\sqrt{{{(4)}^{2}}+{{(2)}^{2}}\times {{10}^{-7}}}\] \[{{B}_{2}}=3\times {{10}^{-4}}T\] The direction of B^ will be perpendecular to S, then, the magnitude of the resultant magnetic induction \[B=\sqrt{B_{1}^{2}+B_{2}^{2}}\] \[=\sqrt{{{(4)}^{2}}+{{(3)}^{2}}\times {{10}^{-4}}}\] \[=5\times {{10}^{-4}}T\]